= Solution
If $|\lambda|>\|T\|$, write $T-\lambda I=-\lambda(I-T/\lambda)$. The <Neumann series> applies to $T/\lambda$, so $\lambda$ is outside the <spectrum of a bounded operator>. Thus $\sigma(T)\subseteq\{\lambda:|\lambda|\leq\|T\|\}$ and $r(T)\leq\|T\|$.
The polynomial <spectral mapping theorem> gives $\sigma(T^m)=\{\lambda^m:\lambda\in\sigma(T)\}$, and hence
$$
r(T)^m=r(T^m)\leq\|T^m\|.
$$
Taking $m$th roots and then the limit, whose existence is allowed, proves
$$
\boxed{r(T)\leq\lim_{m\to\infty}\|T^m\|^{1/m}.}
$$
Only this inequality is required; the reverse inequality is the additional content of the full <spectral radius formula>.
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