What is $\pmatrix{m+n\cr n}$ asymptotically? What is $\pmatrix{m+n\cr n}$ as a function of $n$ asymptotically?
Found a formula that says $\pmatrix{m+n\cr n}\le (e\frac{m+n}{n})^n\sim e^n$ but not sure if the inequality is tight.
 A: Assuming that $m$ is fixed we have for $n\gg m$ by definition of binomial coefficient:
$$
\binom{m+n}{n}\approx\frac{n^m}{m!}.
$$
A: This is very easy to determine. First, $$\binom{n+m}m=\frac{(n+m)(n+m-1)\dots(n+1)}{m!}.$$And it's clear that  $$n+m\sim n,$$
$$n+m-1\sim n,$$etc. So (for $m$ fixed)$$\binom{n+m}m\sim\frac{n^m}{m!}.$$
By the way, looking at your post it appears that you think that $$\left(\frac{n+m}n\right)^n\sim 1.$$That's not right. In fact (again assuming $m$ is fixed)
$$\left(\frac{n+m}n\right)^n=\left(\left(1+\frac mn\right)^{n/m}\right)^m\to e^m.$$
Edit: One can easily derive the more precise asymptotics mentioned by Raymond Manzoni in a comment (note his first comment about that was corrected):
Observe that if $0<k<n$ then $$\log\left(1+\frac kn\right)=\frac kn+O\left(\frac{k^2}{n^2}\right).$$Hence if $n>m$ we have $$\log\left(\left(1+\frac mn\right)\dots\left(1+\frac 1n\right)\right)=\frac{m(m+1)}{2n}+O\left(\frac{m^3}{n^2}\right),$$so$$\begin{align}\frac{\binom{n+m}m}{n^m/m!}
&=\left(1+\frac mn\right)\dots\left(1+\frac 1n\right)
\\&=\exp\left(\frac{m(m+1)}{2n}+O\left(\frac{m^3}{n^2}\right)\right)
\\&=1+\frac{m(m+1)}{2n}+O\left(\left(\frac{m(m+1)}{2n}\right)^2\right)+O\left(\frac{m^3}{n^2}\right)
\\&=1+\frac{m(m+1)}{2n}+O\left(\frac{m^4}{n^2}\right).\end{align}$$
(If you want more terms, just start with more terms in $\log(1+t)=t+\dots$.)
A: Your result does not look correct.
Using Stirling's approximation for $\ln(n!)$ you can show that
$$\ln \binom{n+m}{n} = \ln[(n+m)!]- \ln(n!)-\ln(m!) \sim m \ln n + (m+1) [1- \ln(m+1)] -\frac12 \ln[2\pi/(m+1)] + O(1/n) \,. $$
Thus, we obtain the result
$$ \binom{n+m}{n} = n^m  (m+1)^{-m-1}  e^{m+1} \sqrt{\frac{m+1}{2\pi}} [1 + O(n^{-1}) ]$$
