I have the following function : $$f(k)={n\choose k}(1+e^{-k})^n$$ Here $k$ can vary from $0$ to $n$. I plotted this function for different $n$ on wolfram alpha, like here (for n=20), an found that this function attains maximum value at $k=0$, but I am not able to prove it formally.
Can somebody give a hint how to prove it formally?
Hint : compute the discrete derivative $$ f(k+1)-f(k)= \frac{n!}{(n-k-1)!k!}\bigg(\frac{(1+e^{-k-1})^n}{k+1} -\frac{(1+e^{-k})^n}{n-k}\bigg) $$ and you can then study the function $$ x\mapsto \frac{(1+e^{-x-1})^n}{x+1}-\frac{(1+e^{-x})^n}{n-x}$$ to determine the sign of it (when $x\in[0,n)$).