discrete mathematics - Show that the number $2^{2^n}+1, forall nin{mathbb{N}:ngeq 2}$ has the last digit $7$.

Problem: Show that the number $2^{2^n}+1, \ \forall n\in\{\mathbb{N}:n\geq 2\}$ has the last digit $7$ by induction and without using modular arithmetic.

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I know how induction works and I've tried the basecase $n=2$ which holds true.

Setting $p(n)=2^{2^n}+1$ and computing $p(2)=17,$ which indeed has the last digit seven.

Now I need the inductive step, so I assume this holds true for a number $n=m\in\{\mathbb{N}:m>2\}.$ Our goal here is to verify that $p$ also holds for the following integer $n=m+1.$ Our induction hypothesis states that

$$p(m)=2^{2^m}+1=? \quad \quad (1)$$

Here is where I run into trouble. If the problem statement was to show that it's divisible by $7$, then in $(1)$ I could have set the RHS to $7k$ for some $k\in\mathbb{N}$ and proceed. But in this case, I'm lost.