maximum likelihood of binomial distribution over number of outcomes (k) with known probability (p) I've tried to find the answer to this question, but have been unsuccessful
I'm curious if there is a way to use a ML estimator to find the optimal number of outcomes ($k$) to maximize the binomial distribution:
$$\mathscr{L}(k) = \binom{n}{k}p^{k}(1-p)^{n-k}$$
That is, we know the variables $p$ and $n$. 
Graphing this likelihood for p=0.95 and n=100 gives this:

Obviously, the ML solution should be somewhere around $k = int(np)$, however I'm having trouble "proving" this mathematically.
I have attempted a few solutions, each time the math seems to get away from me. The Solve function in mathematica seems to give an infinite number of solutions (or just overtly incorrect solutions for the Maximize)
In:  Solve[D[Binomial[100, k] 0.95^k 0.05^(100 - k), k] == 0, k]
Out: {{}}
In:  Maximize[Binomial[100, k] 0.95^k 0.05^(100 - k), k \[Element] Integers]
Out: {2.06545*10^-86, {k -> 19}}

 A: Since this is a discrete function with respect to $k$, and there are combinatorial parts in the equation consisting of $k$ again, approaching the problem via derivatives seems not viable. What we are trying to maximize is actually the binomial distribution's PMF. And, there maybe more than one value that maximizes this PMF. For example, if $n=1, p=1/2$; both $k=0,1$ maximizes it, i.e. $\mathscr{L}(0)=\mathscr{L}(1)=1/2$. This isn't the only example, but the simplest one. However, there aren't more than two maxima. The discrete function is going to increase and then decrease. In some situations it'll stay the same at the top before the decrease. Let's examine how this function behaves, as $k$ increases:
For small $k$ it'll increase first, let's see how far it goes:
$$\begin{align}\mathscr{L}(k)\leq\mathscr{L}(k+1) &\rightarrow {n \choose k}p^k(1-p)^{n-k}\leq{n \choose {k+1}}p^{k+1}(1-p)^{n-k-1} \\ &\rightarrow (k+1)(1-p)\leq(n-k)p \rightarrow k\leq p(n+1)-1\end{align}$$
Here, if $k=p(n+1)-1$, then $\mathscr{L}(k)=\mathscr{L}(k+1)$. Typically, this isn't the case since RHS is not an integer. But, for the above example, i.e. $n=0,p=1/2$, RHS becomes an integer, and $k=0$ and $k=1$ represent the maxima. This inequality also tells us that if $k>p(n+1)-1$, then $\mathscr{L}(k)$ is decreasing, meaning we don't have another peak.
Therefore, if $p(n+1)$ is not an integer, the maximizer $k$ will be the integer just above $p(n+1)-1$, which is $\lceil p(n+1)-1 \rceil$ or $\lfloor p(n+1) \rfloor$. And, if it is an integer, there are two maxima: $p(n+1)$ and $p(n+1)-1$. I think there is no need to discuss the degenerate cases, i.e. $p=0$ and $p=1$, since the answers are obvious.
