Prove for $\gcd$ of two integers $a,b$, that $(a^n,b^n) = (a,b)^n$

Let $p_1,p_2,\ldots$ be the increasing enumeration of the primes and let $\prod_{i\ge 1}p_i^{\alpha_i}$ and $\prod_{i\ge 1} p_i^{\beta_i}$ be the prime factorizations of $a$ and $b$, respectively; here $\alpha_i, \beta_i$ are nonnegative integers. Then $$\begin{align} \mathrm{gcd}(a^n,b^n)&=\mathrm{gcd}\left(\prod_{i\ge 1}p_i^{n\alpha_i},\prod_{i\ge 1}p_i^{n\beta_i}\right) \\ &=\prod_{i\ge 1}p_i^{n\min(\alpha_i,\beta_i)} \end{align}$$ and $$\begin{align} \mathrm{gcd}(a,b)^n&=\left(\prod_{i\ge 1}p_i^{\min(\alpha_i,\beta_i)}\right)^n \\ &=\prod_{i\ge 1}p_i^{n\min(\alpha_i,\beta_i)}. \end{align}$$

I always like to prove these sorts of things with Bézout's identity.

In general, if $d\mid e$ then $d^n\mid e^n$.

Letting $d=(a,b)$ and $e=a$ means that $(a,b)^n\mid a^n.$ Likewise, $(a,b)^n\mid b^n,$ so $(a,b)^n\mid (a^n,b^n).$

Now, we prove by induction that we can solve $a^nX+b^nY=(a,b)^n$ for any $n\geq 1.$

We can solve it for $n=1$ by Bézout's identity.

Assume we have $a^nX+b^nY=(a,b)^n.$ Cubing, we get: $$\begin{align}(a,b)^{3n}&=(a^nX+b^nY)^3\\ &=a^{n+1}\left(a^{2n-1}X^{3}+3a^{n-1}b^nX^2Y\right)+b^{n+1}\left(3a^nb^{n-1}XY^2+b^{2n-1}Y^3\right)\\&=a^{n+1}U+b^{n+1}V \end{align}$$

Now notice that both $U$ and $V$ are both divisible by $(a,b)^{2n-1}.$ (This is because $a^{2n-1},a^{n-1}b^n,a^nb^{n-1},$ and $b^{2n-1}$ are.)

So letting $X'=\frac{U}{(a,b)^{2n-1}}, Y'=\frac{V}{(a,b)^{2n-1}},$ we get:

$$a^{n+1}X'+b^{n+1}Y'=(a,b)^{n+1}$$

From these two we conclude that $(a,b)^n\mid(a^n,b^n)$ and $(a^n,b^n)\mid (a,b)^n.$