Prove that every proper subgroup of the group of all $2^n$-th roots of unity is finite.

Think of a pizza. Cut it in half. Those endpoints of the diameter are the square roots of unity. Cut the halves in half. Those endpoints are the additional (primitive) $4$th roots of unity, although the previous points are also $4$th roots of unity. Cut those $4$ pieces in half to produce $4$ additional (again, primitive) $8$th roots of unity, but all the previous $4$ were also $8$th roots of unity. Et cetera.

In summary, all square roots of unity are $4$th roots of unity, which are themselves $8$th roots of unity, etc.

Why? If $z^2 = 1$, then $z^4 = (z^2)^2 = 1^2 = 1$, and $z^8 = (z^4)^2 = 1^2 = 1$, etc.

In general, if $p>n$, then a root of unity $z$ of order $2^n$ is automatically a root of unity of order $2^p$ by the same calculation: $$ z^{2^p} = \bigl( z^{2^n} \bigr)^{2^{p-n}} = 1^{2^{p-n}} = 1, $$ which is meaningful because $p-n>0$.

Aside. This fact is even more general (but not useful in this particular exercise): If $m$ and $n$ are naturals and $m \mid n$, then an $m$th root of unity is automatically an $n$th root of unity. Why? Write $n = mk$ for some $k \in \mathbb{N}$. Then, $$ z^n = z^{mk} = \bigl( z^m \bigr)^k = 1^k = 1. $$