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Why $g(x^{p})=(g(x))^{p}$ in the reduction mod $p$?
Let $h(x) \in \mathbb{Z}[x]$ and $p$ be a prime. We know that for any integer $\alpha$ we have that $\alpha^p \equiv \alpha \pmod{p}$. How can we use this to show that $h(x^p) \equiv h(x)^p \pmod{p}$? It seems to me that we have to reduce the indeterminate $x$ modulo $p,$ which does not make sense.
Imagine expanding $$\left(a_n x^n +a_{n-1}x^{n-1} +\cdots +a_0\right)^p$$ by using the Multinomial Theorem. All of the "mixed" multinomial coefficients are divisible by $p$, so modulo $p$ we end up with $$a_n^p (x^n)^p +a_{n-1}^p(x^{n-1})^p +\cdots + a_0^p,$$ which, modulo $p$, is the same polymomial as $$a_n(x^p)^n +a_{n-1}(x^p)^{n-1}+\cdots +a_0.$$
By the binomial theorem, $$ \begin{align} (ax+by)^p &=\sum_{k=0}^p\binom{p}{k}(ax)^{p-k}(by)^k\\ &\equiv ax^p+by^p\pmod{p}\tag{1} \end{align} $$ because $\left.p\;\middle|\;\binom{p}{k}\right.$ except when $k=0$ or $k=p$, and because $a^p=a\pmod{p}$ for all $a\in\mathbb{Z}$.
Inductively using $(1)$, we get that $$ \left(\sum_{k=0}^na_kx^k\right)^p\equiv\sum_{k=0}^na_kx^{kp}\pmod{p}\tag{2} $$
Hint $ $ By Frobenius / Freshman's Dream $\, z\to z^{p} $ is a ring hom, so commute with polynomials, being compositions of basic ring operations (which, by definition, hom's commute with).
