Understanding a solution that $f(x) = x^n$ is continuous.

I know how to prove that $f(x) = x^n$ is continuous for every $n \in \mathbb{N}$, but there is a particular proof of it I am not able to follow. This is the proof.

Fix $x$ and $\epsilon > 0$. Let $\delta = \frac{\min(\epsilon, 1)}{\max\left(1, \sum\limits_{m=0}^{n-1} \binom{n}{m} |x^m| \right)}$. Fix $h \in (-\delta, \delta)$. Then \begin{align*} |f(x + h) - f(x)| & = |(x+h)^n - x^n| \\ & = \left \lvert \sum\limits_{m=0}^{n-1} \binom{n}{m} h^{n-m} \right \rvert \\ & \leq |h| \sum\limits_{m=0}^{n-1} \binom{n}{m} |x^m h^{n-m}| \\ & \leq |h| \sum\limits_{m=0}^{n-1} \binom{n}{m} |x^m| < \epsilon. \end{align*} Thus, $f$ is continuous.

This proof is very elegant but I can't follow it. The choice of $\delta$ surely came about by scratchwork, but I can't work backwards toward it. The choice of $h$ doesn't make sense to me either, since to show continuity at $x$, I'd want to consider $x'$ satisfying $|x' - x| < \delta$, so $x' \in (x + \delta, x - \delta)$. Unless $x = 0$, it doesn't make sense to take $h \in (- \delta, \delta)$, unless this is captured by considering a small increment in the function values. I assume this must just be an equivalent notion of continuity.

The next few lines seem to use the binomial theorem to rewrite $(x+h)^n$ (I'm not sure what happens to $x^n$) followed by the triangle inequality.

Any thoughts on how to understand this solution would be greatly appreciated.

As aschepler commented in your question, the variable $h$ substitutes $x'-x$ so the continuity is written as:

$$|h| < \delta \implies |f(x+h)-f(x)| < \epsilon$$

of course that:

$$|f(x+h) - f(x)| = |(x+h)^n - x^n| = |\sum_{m=0}^{n-1} {n \choose m} x^m h^{n-m}|$$

we now apply the triangular inequality:

$$|\sum_{m=0}^{n-1} {n \choose m} x^m h^{n-m}| \leq \sum_{m=0}^{n-1} {n \choose m} |x^m h^{n-m}|$$

notice now that the functions $\max$ and $\min$ in the $\delta$ garantee that for any $\epsilon > 0$, for any $x$ and for any $n$ we have $\delta < 1$, thus $0 < |h| \leq 1$ which is not a problem, because it only matters to find the small values of $h$ (if $h_0$ works for some $\epsilon$, then $0<h_1 < h_0$ also works for the same $\epsilon$, right?)

Since $0 <|h| \leq 1$, we can't say that $\sum_{m=0}^{n-1} {n \choose m} |x^m h^{n-m}| \leq |h| \sum_{m=0}^{n-1} {n \choose m} |x^m h^{n-m}|$. Honestly I think this is a typo. Read below.

Now that we are aware that $|h| \leq1$, then $|h|^{n-m} \leq |h|$ simply because $n-m$ is a positive integer ($(\frac 12)^n \leq \frac12$, right?).

So we can use the inequality $|h|^{n-m} \leq |h|$ in every power of $h$ inside the summation:

$ \sum_{m=0}^{n-1} {n \choose m} |x^m h^{n-m}| \leq \sum_{m=0}^{n-1} {n \choose m} |x^m h| = |h| \sum_{m=0}^{n-1} {n \choose m} |x^m|$

Now it is just application of the definition $|h| < \delta \leq \frac{\epsilon}{\sum_{m=0}^{n-1} {n \choose m} |x^m|}$ (late inequality is due the min and max functions)

Notice we can simplify a bit this expression, as $\sum_{m=0}^{n-1} {n \choose m} |x^m| = (|x|+1)^n-1$.

So for the binomial part what they did was this : From the binomial theorem we have that $$|(x+h)^n -x^n| = \left| \left[\sum_{m=0}^{n} {n \choose m}x^{m}h^{n-m}\right] -x^n\right|$$ Taking out the last term we have :

$$|(x+h)^n -x^n| = \left[\sum_{m=0}^{n-1} {n \choose m}x^{m}h^{n-m}\right] +x^n-x^n| = \left| \left[\sum_{m=0}^{n-1} {n \choose m}x^{m}h^{n-m}\right]\right| $$

They seem to not have the $x$ term present anymore, but I feel as if this is a mistake. Do you see where to go from here?