I have an example that derives the MLE for Binomial.
Since there's that factorial term $n_i \choose x_i$ in front of the Binomial p.d.f. and it's a constant, the example claims that one can merely use
$$L(p;x)=\prod_{i=0}^n p^{x_i} (1-p)^{n_i - x_i}$$
as the likelihood function (to find MLE), discarding the constant factorial from it.
Muliplication by a positive constant doesn't affect the MLE. A negative constant will require you to minimize instead of maximize.
Given a function $f$, and a monotonically increasing function $g$, it's not hard to show that $\arg \max g \circ f$ is the same as $\arg \max f$.
For a positive constant $c$, let $g_c(x)=cx$. Obviously $g_c$ is monotonically increasing. In this case, just let $c$ be the reciprocal of whatever constant you'd like to drop from the likelihood.
This is the same reason we can maximize the log-likelihood in place of the likelihood: the logarithm is a monotone function.
The binomial coefficient only varies in $n$ and $x$, but not in $p$, therefore as a function of $p$, the binomial coefficient is constant. Multiplying a function by a positive constant doesn't change the location of the maximum, only its value. MLE only cares about the location of the maximum.
If you use R, you can visualize the MLE manually by plotting the density of the binomial distribution at any given value of the parameter $\pi$, for example in the case of $7$ successes out of $10$ trials (curve in blue on the plot) with this code:
p = seq(0,1,0.0001)
trials= 10
successes = 7
density = dbinom(7, 10, prob=p)
density = (factorial(trials)/(factorial(successes)*factorial(trials-successes))) * p^successes * (1 - p)^(trials-successes)
If now you overlap the plot calculated by multiplying density times 5, the same global maximum will be obtained at $0.7$ (in red).
So, notice that as long as $l(\pi)=\pi^{\text{successes}}\,(1-\pi)^{\text{failures}}$ remains untouched, ${\text{trials}}\choose{\text{failures}}$ in front of it would result in exactly the same estimated $\pi$ with a zero first derivative as any other possible constant.
