Explain why $2^n \mid 2^{2^n}$.
We first note that
$$\begin{align} 2^n & = \underbrace{2 \times 2 \times \ldots \times 2}_{n \text{ prime factors 2}} \\ \\ 2^{2^n} & = \underbrace{2 \times 2 \times \ldots \times 2}_{2^n \text{ prime factor 2}} \end{align}$$
Both $2^n$ and $2^{2^n}$ consist of only of prime factors 2.
To show that $2^n \mid 2^{2^n}$ we need to show that the number of prime factors $2^n$ is more than or equal to $n$.
We will show $2^n \ge n$ by induction.
Base case $n=0$
$2^n = 2^0 = 1 \ge 0 = n$, and so $2^n \ge n$ is true for the base case $n=0$.
Induction Step $2^n \ge n \implies 2^{n+1} \ge n+1$
The induction hypothesis is that $2^n \ge n$, and we aim to show $2^{n+1} \ge n+1$.
$$ \begin{align} 2^{n+1} & = 2(2^n) \\ \\ & \ge 2 (n) \quad \text{induction hypothesis} \\ \\ & = n + n \\ \\ & \ge n + 1 \quad \text{for } n>0 \end{align} $$
We have shown $2^n \ge n \implies 2^{n+1} \ge n+1$. The condition $n>0$ is valid as $n=0$ was handled in the base case.
By induction we have shown that the number of prime factors 2 in $2^{2^n}$ is greater than or equal to the number of prime factors 2 in $2^n$, and so $2^n \mid 2^{2^n}$.