Can someone suggest a way to calculate the maximum with respect to $x \ge 1$ of:
$$f(x)=\frac{1}{x!} \frac{1}{1-c^{1/\binom{x+n-1}{n-1}}}.$$
The constants $c$ and $n$ are parameters such that $c \in (0,1)$ and $n \in \mathbb{N}$.
If it helps, we can assume $x \in \mathbb{N}$ but it is not necessary.  

I've tried looking at f(x+1)/f(x) which does simplifies things enough to solve it. 
Working it out a bit, I suspect the maximum is for $x$ such that $\lfloor\sqrt{n} \rfloor -1 \le x \le \lfloor\sqrt{n} \rfloor$. 



maximizing a function involving factorial
In a quite large range of the parameters we can approximate the fraction by a Taylor series to obtain
$$
f(x) = \frac{1}{x!}\frac{1}{\frac{-\log c}{\binom{x+n-1}{n-1}} + \mathcal{O}\left(\frac{\log^2 c}{\binom{x+n-1}{n-1}^2}\right)} 
= \left(1+ \mathcal{O}\left(\frac{\log^2 c}{\binom{x+n-1}{n-1}^2}\right)\right)\frac{\binom{x+n-1}{n-1}}{-x!\log c},
$$
which is valid for $-\log c<\binom{x+n-1}{n-1}$. Hence the function to be optimized is essentially $g(x)=\frac{1}{x!}\binom{x+n-1}{n-1}$. Now $\frac{g(x+1)}{g(x)}=\frac{x+n}{x(x+1)}$, hence the maximum is attained around $x=\sqrt{n}$. If $x$ differs from this optimum by more than 1, then $\frac{g(x+1)}{g(x)}$ differs from 1 by more than $\frac{1}{x}$, hence the error term in our formula for $f$ is negligible provided that $-\log c$ is small compared to $n^{\sqrt{n}}$. Whether this is enough depends on what your parameters evetually are. If $c$ is much closer to 0, then the maximum will be at much smaller values of $x$.


