I have this function I'm trying to find the maximum of by using Stirling's approximation for factorials
$$ \frac{{N \choose 2}!}{m! \left( {N \choose 2} - m \right)!} \approx \frac{\Gamma({N \choose 2}+1)}{\Gamma(m+1)\Gamma({N \choose 2}-m+1)} $$
where $$ \Gamma'(m+1) = m!\left( - \gamma \sum_{k=1}^m \frac{1}{k} \right) $$ for positive integers
My problem is when I calculate the derivative I get a messy equation that I'm having trouble solving. Is there a better way to cancel out this function or take the derivative?
As I mentioned in the comments, you don't need to know anything about derivatives or Gamma functions or even Stirling's approximation to do this. The following is totally elementary:
Proof. We just calculate that
$$\frac{ {N \choose m+1} }{ {N \choose m} } = \frac{m!(N-m)!}{(m+1)!(N-(m+1))!} = \frac{N - m}{m+1}.$$
From this it follows immediately that ${N \choose m+1} \ge {N \choose m}$ if and only if $\frac{N - m}{m + 1} \ge 1$ if and only if $N \ge 2m+1$, and similarly for strict inequalities. This gives that the function $m \mapsto {N \choose m}$ is strictly increasing for $0 \le m \le \frac{N-2}{2}$ and strictly decreasing for $\frac{N}{2} \le m \le N$.
If $N$ is even we get that the maximum occurs when $m = \frac{N}{2}$ and if $N$ is odd we get that the maximum occurs at both $m = \left\lfloor \frac{N}{2} \right\rfloor$ and $m = \left\lceil \frac{N}{2} \right\rceil$. $\Box$
Heuristically this happens because by the central limit theorem a binomial distribution approaches a normal distribution centered at its mean, which is $\frac{N}{2}$. It may help to stare at a graph of the probability mass function of a binomial distribution, which is given at the Wikipedia article. The above calculation can be adapted to finding the maximum of ${N \choose m} p^m (1 - p)^{N-m}$ as well (a binomial distribution with parameter $p$, corresponding to $N$ flips of a coin with bias $p$), which as expected occurs near the mean $Np$.