= Solution
The <Neumann series> criterion states that $\|A\|<1$ for a <bounded linear operator> on a <Banach space> implies
$$
(I-A)^{-1}=\sum_{n=0}^{\infty}A^n,
$$
with convergence in the <operator norm>. In particular, $\|T\|<1$ suffices for invertibility of $I-T$.
If $|\lambda|>\|T\|$, apply this criterion to $A=T/\lambda$. It gives
$$
(\lambda I-T)^{-1}=\lambda^{-1}\sum_{n=0}^{\infty}(T/\lambda)^n.
$$
Thus no such $\lambda$ belongs to the <spectrum>, and
$$
\boxed{r(T)\leq\|T\|.}
$$
For each integer $j\geq1$, if $|\lambda|^j>\|T^j\|$, then $\lambda^jI-T^j$ is invertible by the same argument. The commuting factorization
$$
\lambda^jI-T^j=(\lambda I-T)\sum_{k=0}^{j-1}\lambda^{j-1-k}T^k
$$
then gives a bounded two-sided inverse for $\lambda I-T$, namely the displayed sum multiplied by $(\lambda^jI-T^j)^{-1}$. Hence every $\lambda\in\sigma(T)$ satisfies $|\lambda|^j\leq\|T^j\|$. Taking the <supremum> over the <spectrum> and then the <infimum> over $j$ proves
$$
\boxed{r(T)\leq\inf_{j\geq1}\|T^j\|^{1/j}.}
$$
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