<s>[INST] $(a^{n},b^{n})=(a,b)^{n}$ and $[a^{n},b^{n}]=[a,b]^{n}$? How to show that $$(a^{n},b^{n})=(a,b)^{n}$$ and $$[a^{n},b^{n}]=[a,b]^{n}$$ without using modular arithmetic? Seems to have very interesting applications.$$$$Try: $(a^{n},b^{n})=d\Longrightarrow d\mid a^{n}$ and $d\mid b^n$
 [/INST] Bezout's Identity says that $\gcd(a,b)$ is the smallest positive element of $\{ax+by:x,y\in\mathbb{Z}\}$.
The smallest positive element of $\{acx+bcy:x,y\in\mathbb{Z}\}$ is $\gcd(ac,bc)$; it is also $c$ times the smallest positive element of $\{ax+by:x,y\in\mathbb{Z}\}$ which is $\gcd(a,b)$. Therefore
$$
\gcd(ac,bc)=c\gcd(a,b)\tag{1}
$$
Suppose $ax+by=1$ and $au+cv=1$, then
$$
\begin{align}
by\,cv&=(1-ax)(1-au)\\
&=1-a(x+u-axu)\\
a(x+u-axu)+bc\,vy&=1
\end{align}
$$
Thus,
$$
\gcd(a,b)=1\quad\text{and}\quad\gcd(a,c)=1\implies\gcd(a,bc)=1\tag{2}
$$
Using $(2)$ and induction, we get that
$$
\gcd(a,b)=1\implies\gcd\left(a^n,b^n\right)=1\tag{3}
$$
Using $(1)$ and $(3)$, we have
$$
\begin{align}
\gcd(a,b)\gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)&=\gcd(a,b)&&(1)\\
\gcd\left(\frac{a}{\gcd(a,b)},\frac{b}{\gcd(a,b)}\right)&=1&&\text{cancel}\\
\gcd\left(\frac{a^n}{\gcd(a,b)^n},\frac{b^n}{\gcd(a,b)^n}\right)&=1&&(3)\\
\gcd(a,b)^n\gcd\left(\frac{a^n}{\gcd(a,b)^n},\frac{b^n}{\gcd(a,b)^n}\right)&=\gcd(a,b)^n&&\text{multiply}\\[9pt]
\gcd\left(a^n,b^n\right)&=\gcd(a,b)^n&&(1)\tag{4}
\end{align}
$$

In this answer, it is shown that
$$
\gcd(a,b)\,\mathrm{lcm}(a,b)=ab\tag{5}
$$
Therefore,
$$
\begin{align}
\mathrm{lcm}(a,b)^n
&=\frac{(ab)^n}{\gcd(a,b)^n}&&(5)\\
&=\frac{a^n\,b^n}{\gcd(a^n,b^n)}&&(4)\\[4pt]
&=\mathrm{lcm}(a^n,b^n)&&(5)\tag{6}
\end{align}
$$
</s>