MLE estimator for the second parameter of binominal distribution Let us have a sample $X_1, X_2, ... , X_n$ with $B_{p,m}$ distribution.
How to make an estimator for $m$ using MLE ? For simplicity, let us $n=1$
I am a little bit stuck, because I have a derivative of binominal coefficient which is uncommon in usual undergrad textbooks.. So, probably, I am doing something wrong.
Update: $p$ is known
 A: I don't see how to do this with only one observation, if $p$ is unknown.
If $p$ is known, here are some clues. [More generally, this is a much-studied problem; perhaps see this paper and its references.]
If $p = 0.3$ and your observation is $X = 12,$ then
the method of moments estimator is $X/p = 40.$
Perhaps it is reasonable to guess that the MLE will
be approximately the same.
Here is a graphical solution, using R:
m = 1:150; p = .3; x = 12
like = dbinom(x, m, p)
plot(m, like, type="l", lwd=2)
 abline(h=0, col="green2")
mle = mean(m[like==max(like)]);  mle
[1] 39.5
 abline(v=mle, col="red")


A: We have the likelihood of $m$ by $$f(m)=\frac{m!}{(m-X)!X!}p^X(1-p)^{m-X}$$
Here $m\geq X$.
Now, let's consider when $\frac{f(m+1)}{f(m)}$ is less than 1. When it starts to be less than 1, we know that function $f$ reached a (local) maximum. After calculation we find
$$r = \frac{f(m+1)}{f(m)} = \frac{(m+1)(1-p)}{m+1-X},$$ and when $n\geq\left\lfloor\frac{X}{p}\right\rfloor$, we have $r\leq 1$. Thus, we conclude that $$\left\lfloor\frac{X}{p}\right\rfloor\text{ is the MLE of }m.$$
Here, $\lfloor y\rfloor$ denotes the largest integer that is less than $y\in\mathbb{R}.$
----- update -----
By the way, BruceET's numerical example confirms the conclusion.
