TITLE: Prove if $U$ unitary, $U^k$ has convergent subsequence to $I$
QUESTION [5 upvotes]: Let $U$ be a unitary matrix in $M_{n}(\mathbb{C})$. Prove $\{U^{k}\}_{k \in\Bbb N}$ has a subsequence that converges to $I$ (identity matrix). How to prove that?
I tried using the Spectral theorem.

REPLY [2 votes]: Observation 1: The set of unitary matrices is compact.
Proof: The operator norm of a unitary is 1, so the set of unitary matrices is bounded.  The defining relation $U^{\dagger}U = 1$ shows that the set of unitaries is the preimage of a closed set $\{1\}$ of a continuous map, so the set of unitaries is closed.  Closed & bounded implies compact.
Observation 2: A sequence in a compact set has a convergent (Cauchy) subsequence.  Call the limit of the subsequence $L$.
So for any $\epsilon>0$ we can find a $q$ s.t. for any element $U^p$ of the subsequence with $p\geq q$, $||L-U^p||<\epsilon/2$.  WLOG pick $q$ s.t. $U^q$ is itself in the subsequence.  For any $n\in \mathbb{N}$ such that $U^{n+q}$ is in the Cauchy subsequence, $$\epsilon > ||L-U^{n+q}|| + ||L-U^q|| \geq ||U^{n+q}-U^q|| = ||U^q (U^n-1)|| = ||U^n-1||.$$  So there exist arbitrarily high powers of $U$ arbitrarily close to 1.  Any Cauchy subsequence of these must converge to 1. 
No spectral theorem required!
(Thanks to A. Barron's comment on the OP for suggesting this approach.)