$$\frac{N}{m^2}((N+m)\ln(N+m)+(N-m)\ln(N-m)-2N\ln(N))\to 1 \text{ when } N \gg m$$
I got this expression from fiddling around with Brownian motion. From inputing values for $N$ and $m$ I can see that it goes towards 1 very quickly, even with small differences, like $N=10$ and $m=1$. Is L'Hôpital the way to go here or what?
Edit: $\frac{N}{m}$ was wrong, the $m$ should be squared like it is now.
You can rewrite this as
$$ \frac{1}{m^2} \ln{\left( \frac{(N+m)^{N+m}(N-m)^{N-m}}{N^{2N}} \right)^N}$$
Then you want to show that the quantity inside the logarithm converges to $e$. Write it as $$\left( \frac{(N+m)^m}{(N-m)^m}\right)^N \cdot \left( \frac{N^2-m^2}{N^2}\right)^{N^2}$$.
The first part goes to $e^{2m^2}$, the second is $e^{-m^2}$. You can see the second part since it is $$\left(1-\frac{m^2}{N^2}\right)^{N^2}.$$ It is a known calculus fact that $(1+\frac{x}{n})^n \rightarrow e^x$ so you use this. As for the first part, write it as $$e^{mN\ln{\frac{N+m}{N-m}}}$$ and use L'Hospital.