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Using Lagrange interpolation (I think identity $\sum \limits_{k=0}^{n}k^{m}\prod \limits_{\substack{i=0\\i\neq k}}^{n}\frac{x-i}{k-i}=x^m$) shows that

$$\sum \limits_{k=0}^{n}(-1)^{k}k^{m}\binom{n}{k}=0 \text{ if }\ 0≤m

and $$\sum \limits_{k=0}^{n}(-1)^{k}k^{n}\binom{n}{k}=(-1)^{n}n!$$ (same sum with $m=n$).

EDIT: I found it quite straightforward to prove first equality (just put x=0 into identity), the second part is definitely harder.

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    Is Lagrange interpolation a requirement for an acceptable solution here? – 2017-02-24
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    Another solution is also good. – 2017-02-24

3 Answers 3

3

It can be easily proven using partial fraction expansion. Since this expansion is entailed by Lagrange's interpolation formula for the numerator, it may give you an interesting view of the problem.

For $0 \leqslant m \leqslant n$ we have the expansion \begin{equation} \frac{t^{m}}{t(t+1)\dotsm(t+n)}=\sum_{k=0}^{n}{\frac{c_{k}}{t+k}}.\label{eq}\tag{E} \end{equation} You get $c_{k}$ by multiplying by $t+k$ and then evaluate at $-k$, which gives: $$\frac{k^{m}}{(k)(k-1)\dotsm (1)(-1)\dotsm(-(n-k))}=c_{k}$$ that is $c_{k}=(-1)^{n-k}\binom{n}{k}\frac{k^{m}}{n!}$. Multiply the expansion \eqref{eq} by $t$ and have $t \to +\infty$ to get $$\sum_{k=0}^{n}{c_{k}}=\begin{cases}0&\text{if $m

Now, the link with Lagrange's interpolation: if you write the interpolation formula at points $-n,\dotsc,-1,0$ you get: $$t^{m}=\sum_{k=0}^{n}{\left((-k)^{m}\prod_{\substack{i=0\\i \neq k}}^{n}{\frac{t+i}{-k+i}}\right)}$$ Divide by $t(t+1)\dotsm(t+n)$ to recognize the partial fraction expansion above.

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    Very impressive, for me completely fresh point of view. I hope there will be more answers like this. – 2017-02-24
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    @Shingle Thank you. Have a close look to Dr.MV's answer, it's also very interesting and a completly different point of view. – 2017-02-24
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \sum_{k = 0}^{n}\pars{-1}^{k}\,k^{m}{n \choose k} & = \sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}\bracks{% m!\oint_{\verts{z} = 1}{\exp\pars{kz} \over z^{m + 1}}\,{\dd z \over 2\pi\ic}} \\[5mm] & = m!\oint_{\verts{z} = 1}{1 \over z^{m + 1}} \sum_{k = 0}^{n}{n \choose k}\pars{-\expo{z}}^{k}\,{\dd z \over 2\pi\ic} = m!\pars{-1}^{n}\oint_{\verts{z} = 1}{\pars{\expo{z} - 1}^{n} \over z^{m + 1}} \,{\dd z \over 2\pi\ic} \\[5mm] & = m!\pars{-1}^{n}\oint_{\verts{z} = 1}{1 \over z^{m + 1}} \bracks{n!\sum_{k = 0}^{\infty}{k \brace n}{z^{k} \over k!}} \,{\dd z \over 2\pi\ic} = \bbx{\ds{\pars{-1}^{n}\,n!}\,{m \brace n}} \end{align}

$\ds{a \brace b}$ is a Stirling Number of the Second Kind.

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    Very nice. (+1). I used a brute attack. – 2017-02-24
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    Thank you for this other point of view. This one need a little bit more experience in higher mathematics but is very elegant. – 2017-02-24
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    @Dr.MV Thanks. Any attack is fine. – 2017-02-24
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    @TiglaPomme You're welcome !!!. – 2017-02-24
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    I think it will take me some time to understand that (actually I have no idea about complex analysis), but I wrote it down for that moment. Thank for this solution, this is something that I couldn't even imagine to see. – 2017-02-25
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    @Shingle Thanks. It just involves some simple application of Residues Theorem. – 2017-02-25
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I thought it might be instructive to present an alternative approach. Using the binomial theorem, we can write

$$(1-x)^n=\sum_{k=0}^n\binom{n}{k}(-1)^{k}x^k\tag 1$$

Differentiating $(1)$ with respect to $x$ and multiplying the result by $x$ yields

$$\begin{align} x\frac{d}{dx}\left((1-x)^n\right)&=-nx(1-x)^{n-1}\tag2\\\\ &=\sum_{k=0}^n\binom{n}{k}(-1)^{k}kx^k \tag3 \end{align}$$

Setting $x=1$ in $(2)$ and $(3)$ reveals

$$\sum_{k=0}^n\binom{n}{k}(-1)^{k}k=0$$

provided $n>1$. And if $n=1$, then we find that $\sum_{k=0}^1\binom{1}{k}(-1)^kk^1=1$.

We can repeat this process to obtain

$$\left.\left(\left(x\frac{d}{dx}\right)^m (1-x)^n\right)\right|_{x=1}=\sum_{k=0}^n\binom{n}{k}(-1)^{k}k^m \tag 4$$

So, the problem boils down to showing that the left-hand side of $(4)$ is $0$ for $m

We claim that we can write

$$\left(x\frac{d}{dx}\right)^m (1-x)^n=\sum_{\ell=1}^m p_\ell(x)(1-x)^{n-\ell} \tag {5a}$$

where $p_\ell(x)$ is a polynomial of order $\ell$ with

$$p_m(x)=(-1)^m m!\binom{n}{m}x^m \tag {5b}$$

Proving $(5)$ by induction

From $(1)$, we see that a base case to prove $(5)$ inductively is established. Then, assuming $(5)$ is true for some $m>1$, we have

$$\begin{align} \left(x\frac{d}{dx}\right)^{m+1}(1-x)^{n+1}&=x\frac{d}{dx}\sum_{\ell=1}^m p_\ell(x)(1-x)^{n-\ell}\\\\ &=\sum_{\ell=1}^m \left(xp_\ell'(x)(1-x)^{n-\ell}-(n-\ell)xp_\ell(x)(1-x)^{n-\ell-1}\right)\\\\ &=\sum_{\ell=1}^{m+1} q_\ell(x)(1-x)^{n-\ell} \end{align}$$

where the terms $q_\ell(x)$ are given by

$$q_\ell(x)=\begin{cases}xp_\ell'(x)&,\ell=1\\\\ xp_\ell'(x)-(n-\ell+1)xp_{\ell-1}(x)&,1<\ell\le m\\\\ -(n-m)xp_{m}(x)&,\ell=m+1 \end{cases}$$

Furthermore, we see that

$$\begin{align} q_{m+1}(x)&=(-1)(n-m)xp_m(x)\\\\ &=(-1)^{m+1}(n-m)m!\binom{n}{m}\\\\ &=(-1)^{m+1}(m+1)!\binom{n}{m+1} \end{align}$$

And we have proven inductively that the relationships in $(5)$ hold.

Finally, using $(5)$ in $(4) yields

$$\sum_{k=0}^n\binom{n}{k}(-1)^{k}k^m =\begin{cases}0,n>m\\\\(-1)^n\,n!&,n=m\end{cases}$$

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    This approach is more efficient if you differentiate $m$ times the function $t \mapsto (1-e^{t})^{n}$: the derivation of the binomial expansion gives straightforward the sum with $k^{m}$, and the Taylor expansion $(1-e^{t})^{n}=(-1)^{n}t^{n}+o(t^{n})$ shows that $f^{(m)}(0)=0$ for $m$(-1)^{n}n!$ for $m=n$. – 2017-02-24
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    It always surprises me when I see a long, thoughtful answer like this. Keep it up if you can! – 2017-02-24
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    @ChrisDugale Chris, thank you! Much appreciative. -Mark – 2017-02-24
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    @Dr.MV The $m$-th derivative of $\sum_{k=0}^{n}{\binom{n}{k}(-1)^{k}\mathrm{e}^{kt}}$ is straightforward (and gives $\sum_{k=0}^{n}{\binom{n}{k}(-1)^{k}k^{m}\mathrm{e}^{kt}}$), whereas using $\sum_{k=0}^{n}{\binom{n}{k}(-1)^{k}x^{k}}$ you need to multiply by $x$ after each derivative to get the $k^{m}$ factor in the sum, which makes you need the $(5)$ formula (which you need to prove apart of the initial question). – 2017-02-24
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    @TiglaPomme Yes, I suppose that streamlines the right-hand side. But the real boost in efficiency comes from the Taylor expansion of $(1-e^t)^n=(-1)^nt^n+O(t^{n+1})$. I've never seen this relationship before, and so was not equipped with that trick. Where had you seen it? – 2017-02-24
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    @Dr.MV Yes, the deep interest is the Taylor expansion, which hides a $n$-th iterate of a convolution product. It's a big boost, similar to the generating series approach for combinatorics. – 2017-02-24
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    @TiglaPomme Interestingly, the equivalent way forward would be to recognize that for $x$ near $1$, $x-1\sim \log(x)$. Furthermore, $x\frac{d}{dx}\log^n(x)=n\log^{n-1}(x)$ and we now see that $\left(x\frac{d}{dx}\right)^m \log^n(x)=n(n-1)\cdots (n-m+1)\log^{n-m}(x)$. So, the transformation $x\to e^t$ simplifies the cumbersome notation. – 2017-02-24
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    I just readed it in detail, thanks for this answer. I tried applying induction directly, but this is definitely smoother way in such case. – 2017-02-24
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    @shingle You're welcome. My pleasure. Pleased you have it! -Mark – 2017-02-24