Proof of Gelfand's formula without using $\rho(A) < 1$ iff $\lim A^n = 0$

We assume without proof that Gelfand's formula holds in the easy case where $A$ is diagonalisable. For the general case, since all norms are equivalent on a finite-dimensional vector space, it suffices to prove Gelfand's formula using Frobenius norm.

Frobenius norm is unitarily invariant. So, we may assume that $A$ is triangular. Denote by $|A|$ the entrywise absolute value of $A$. Let $\{D_k\}$ be any sequence of nonnegative diagonal matrices such that all diagonal entries of $D_k$ are smaller than $\frac1k$ and all eigenvalues of $|A|+D_k$ (which are its diagonal entries) are distinct.

For every $k$, as $|A|+D_k$ is diagonalisable, Gelfand's formula holds. Therefore there exists some $n_k$ such that $$ \|(|A|+D_k)^n\|_F^{1/n}<\rho(|A|+D_k)+\frac1k\tag{1} $$ whenever $n\ge n_k$. However, as $A$ is triangular and $D_k$ is entrywise bounded above by $\frac1k$, we have $\rho(|A|+D_k)+\frac1k<\rho(A)+\frac2k$. Hence $(1)$ implies that $$ \rho(A)\le\|A^n\|_F^{1/n}\le\|(|A|+D_k)^n\|_F^{1/n}<\rho(A)+\frac2k $$ whenever $n\ge n_k$ and our conclusion follows.