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I can't figure out an algebraic proof for the following identity, (and i don't know if i can use the binomial theorem for this one): $$\sum_{k=0}^m(-1)^{m-k}{{n \choose {k}}}= {{n-1}\choose m}$$

Thank you, for your help in advance.

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    Have you tried writing $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$? – 2017-02-07
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    @ΘΣΦGenSan who knows :/ I posted an answer over 30 minutes ago and haven't received a word from the OP – 2017-02-08

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We desire to prove, by induction, that $$\sum_{k=0}^m (-1)^{k} {n \choose k} = (-1)^m{n-1 \choose m}$$ We start with the case $m=0$ $$\sum_{k=0}^0 (-1)^{k} {n \choose k} = {n \choose 0}= {n-1 \choose 0}$$ We now apply induction. We first assume that $$\sum_{k=0}^{m-1} (-1)^{k} {n \choose k} = (-1)^{m-1}{n-1 \choose m-1} \tag{1}$$ And now we begin the induction $$\sum_{k=0}^{m} (-1)^{k} {n \choose k}$$ $$=(-1)^{m}{n \choose m}+\sum_{k=0}^{m-1} (-1)^{k} {n \choose k}$$ $$=(-1)^{m}{n \choose m}+(-1)^{m-1}{n-1 \choose m-1}$$ $$=(-1)^{m}\left[{n \choose m}-{n-1 \choose m-1}\right]$$ We now have to apply the identity $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$ and we are done.

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    It seems that your claim leads to $$\sum_{k=0}^m (-1)^{k-m} {n \choose k} = {n-1 \choose m}$$ different to OP's question $$\sum_{k=0}^m (-1)^{m-k} {n \choose k} = {n-1 \choose m}$$ – 2017-02-08
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    @ΘΣΦGenSan Note that $(-1)^{k-m} = (-1)^{-(m-k)} = (-1)^{m-k}$. The two are not different – 2017-02-08
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    @ΘΣΦGenSan This comes down to the fact that $(-1)^k = \frac{1}{(-1)^k}$ – 2017-02-08
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    --)Nice trick. I did not see that. Thank you. One up vote from me. – 2017-02-08
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$$ (1+x)^{n-1} = \sum_{m=0}^n \binom{n-1}m x^m $$ but also $$ \begin{align} (1+x)^{n-1} =(1+x)^n \frac1{1+x} & = \sum_{k=0}^n \binom{n}{k}x^k\sum_{l=0}^{\infty}(-1)^lx^l \\ & =\sum_{m=0}^{\infty}x^m\sum_{k=0}^m (-1)^{m-k}\binom{n}k \tag{2} \end{align} $$ now compare the coefficients of $m$ in the two expansions. although (2) is formally an infinite sum, you can see that for $m \gt n-1$ the terms are zero

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$$\begin{align}\require{cancel} \sum_{k=0}^m(-1)^{m-k}\binom nk &=(-1)^m\sum_{k=0}^m (-1)^k\left[\binom {n-1}{k-1}+\binom {n-1}k\right] &&\text{as $(-1)^{-k}=(-1)^k$}\\ &=(-1)^m\bigg\lbrace \cancel1-\left[\cancel{\binom {n-1}0}+\bcancel{\binom {n-1}1}\right]\\ &\qquad\quad\quad\quad+\left[\bcancel{\binom {n-1}1}+\cancel{\binom {n-1}2}\right]\\ &\qquad\quad\quad\quad-\left[\cancel{\binom {n-1}2}+\bcancel{\binom {n-1}3}\right]\\ &\qquad\qquad\qquad\cdots\\ &\qquad\;\;+(-1)^m\left[\bcancel{\binom {n-1}{m-1}}+\binom {n-1}m\right]\bigg\rbrace\\ &=\binom {n-1}m\end{align}$$