Cayley-Hamilton theorem can be used to prove Gelfand's formula (whose usual proofs rely either on complex analysis or normal forms of matrices).
Let $A$ be a $d\times d$ complex matrix, let $\rho(A)$ denote spectral radius of $A$ (i.e., the maximum of the absolute values of its eigenvalues), and let $\|A\|$ denote the norm of $A$. (Fix your favorite matrix norm.)
Gelfand's formula: $\rho(A) = \lim_{n \to \infty} \|A^n\|^{1/n}$.
Proof: The choice of the norm does not affect the validity of the result, so choose an operator norm. The existence of the limit $r:=\lim_{n \to \infty} \|A^n\|^{1/n}$ is a simple consequence of submultiplicativity of the norms (by Fekete Lemma). Existence of (complex) eigenvectors implies $r \ge \rho(A)$, so the nontrivial part is to show that $r \le \rho(A)$.
It follows from Cayley-Hamilton theorem that $$\|A^d\|\le C \rho(A) \|A\|^{d-1},$$ where $C>0$ is a constant that depends only on $d$; this is a straightforward exercise -- or see this post by Ian Morris for the details.
Applying this inequality to $A^n$ and taking the $n$-th root we obtain: $$\|A^{dn}\|^{1/n}\le C^{1/n} \rho(A) \|A^n\|^{(d-1)/n}.$$ Making $n \to \infty$ we get $r^d \le \rho(A) r^{d-1}$, thus concluding the proof of Gelfand's formula.
Reference:
Jairo Bochi, Inequalities for numerical invariants of sets of matrices, Linear Algebra Appl. 368 (2003), 71--81.