<s>[INST] Asymptotics for $k^c +k - 1 \choose k$ when $c\geq 1$ 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 $${k^c +k - 1 \choose k} \sim \frac{k^{ck-1}e^k}{\sqrt{2 \pi k}}\;?$$
 [/INST] 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 $1<c<2$, $$\binom{k^c+k-1}{k}\sim\frac{(k^c+k-1)^k}{k!}\exp\left(-\frac{k^{2-c}}2\right)\sim\frac{k^{ck}}{k!}\exp\left(+\frac{k^{2-c}}2\right)$$
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