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I am looking for the asymptotics of $n +k - 1 \choose k$ when $n = k^c$ for integer constant $c\geq 1$.

From this we get:

$$\frac{(n+k-1)!}{k! (n-1)!} \approx \frac{\sqrt{2 \pi(n+k-1)}\left(\frac{n+k-1}{e}\right)^{n+k-1}}{\sqrt{2 \pi k}\left(\frac{k}{e}\right)^k \sqrt{2\pi(n-1)}\left(\frac{n-1}{e}\right)^{n-1}} = \frac{\sqrt{(n+k-1)}\left(n+k-1\right)^{n+k-1}}{\sqrt{ k}\left(k\right)^k \sqrt{2\pi(n-1)}\left(n-1\right)^{n-1}}$$

We also have that,

$$ \frac{(n+k-1)!}{k! (n-1)!} = \frac{(k^c+k-1)!}{k! (k^c-1)!} $$ I am not sure where to go from here.

Blindly copying from the wiki and assuming that "$k$ is much smaller than $n$" is satisfied, you get:

$$ {k^c +k - 1 \choose k} \approx \frac{(\frac{k^c+k-1}{k} - \frac{1}{2})^k e^k}{\sqrt{2 \pi k}} $$

Does this mean that $${k^c +k - 1 \choose k} \sim \frac{k^{ck-1}e^k}{\sqrt{2 \pi k}}\;?$$

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    [This](https://en.wikipedia.org/wiki/Binomial_coefficient#Bounds_and_asymptotic_formulas) may interest you. – 2017-02-09
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    @S.C.B. I gave it my best go using the wiki info but I am not at all sure about the result. – 2017-02-09
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    Since $c$ is an *integer* constant, it will be very useful to consider the case $c=1$ separately from $c\ge 2$, where indeed $k$ is much smaller than $n$. – 2017-02-09
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    @felipa: Your final asymptotic formula is correct, except for a typo in the exponent of $k $. See my answer below. – 2017-02-09
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    The link provided by S.C.B. answers the case $c=1$, once you realize that $\binom{2k-1}{k} = \tfrac12 \binom{2k}{k}$. – 2017-02-10

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Note that $$\frac{k!}{(k^c+k-1)^k}\binom{k^c+k-1}{k}=\prod_{i=1}^k\frac{k^c+i-1}{k^c+k-1}=\prod_{i=1}^{k-1}\left(1-\frac{i}{k^c+k-1}\right)$$ Since $c>1$, the last product is asymptotically equivalent to $$\prod_{i=1}^{k-1}\left(1-\frac{i}{k^c}\right)\approx\prod_{i=1}^{k-1}\exp\left(-\frac{i}{k^c}\right)\approx\exp\left(-\frac{k^2}{2k^c}\right)$$ Hence, if $c>2$, $$\binom{k^c+k-1}{k}=\frac{(k^c+k-1)^k}{k!}\left(1+O\left(\frac1{k^{c-2}}\right)\right)$$ while $$\binom{k^2+k-1}{k}\sim\frac{(k^2+k-1)^k}{k!}e^{-1/2}\sim\frac{k^{2k}}{k!}e^{+1/2}$$ and, for every $1

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    What is not already "covered" when $c>2$? – 2017-03-13
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    My mistake. Thank you again! – 2017-03-13
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    I am sorry for not getting this but I think we know that $\binom{k^c+k-1}{k} \leq \frac{(k^c+k-1)^k}{k!}$ and $k$ is positive. If $c >2$, then it seems $O(1/k^{c-2})$ has to equal at most $0$. Is that right? If so, what is the purpose of the term? I am not used to O terms being non-positive. – 2017-03-13
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    Yes, as I already explained, $O$ terms can have both signs, for example $\frac{n-3}n=1+O\left(\frac1n\right)$. – 2017-03-13
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    Oh it refers to the absolute value! Thank you I had no idea. – 2017-03-13
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    I am worried about $1 < c < 3/2$. I think we might need more terms in the expansion. – 2017-03-13
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    I am pretty sure this gives the wrong answer for $c = 4/3$ for example. – 2017-03-13
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    Excellent -- care to explain why? – 2017-03-13
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    I think the answer when $c = 4/3$ is asymptotic to $\frac{1}{\sqrt{2\pi k}}\left(\frac{k^c}{k}\right)^k e^{k+\frac{k^{2/3}}{2} - \frac{k^{1/3}}{6} + \frac{1}{12}}$ – 2017-03-13
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    So... if we retranslate this in terms of $k!$, this reads $$\frac{k^{ck}}{k!}\exp\left(+\frac{k^{2/3}}2\color{red}{-\frac{k^{1/3}}6+\frac1{12}+o(1)}\right)$$ – 2017-03-13
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    Yes. For $c = 3/2$ I think you get $\frac{1}{\sqrt{2\pi k}}\left(\frac{k^c}{k}\right)^k e^{k + \sqrt{k} - 1/6}$ . For $ c> 3/2$ our answers agree. – 2017-03-14