Although this question was answered over 5 years ago, I will add this answer, which gives an alternative explanation from a linear transformation perspective.
Consider a vector-induced matrix norm $\Vert A \Vert = \mathrm{sup}_{\bar{x}\neq 0}\frac{\Vert A\bar{x}\Vert}{\Vert \bar{x} \Vert}$ of an $n\times n$ real matrix $A$ (this explanation also works if $A$ is complex, but I will leave that out for brevity).
Let $\bar{v}_1,\bar{v}_2,...\bar{v}_n $ be the normalized eigenvectors of $A$, and $\lambda_1,\lambda_2,...\lambda_n \in \mathbb R$ be the corresponding eigenvalues (this explanation also works if $\lambda_i$ are complex conjugate pairs, but I will leave that out for brevity).
Any unit vector $\bar{u}$ in the space spanned by $A$ can be given by
$$\bar{u}=(w_1 \bar{v}_1 +w_2\bar{v}_2+...+w_n\bar{v}_n)$$
where, $ w_i\in \mathbb R$ are scaling factors.
Any non-zero vector $\bar{x}$ can be given as $\bar{x}=k\bar{u}$, where $k>0\in \mathbb R$
$\therefore$ $A\bar{x}=k(\lambda_1w_1\bar{v}_1 +\lambda_2w_2\bar{v}_2+...+\lambda_nw_n\bar{v}_n)$
Notice $\Vert \bar{x} \Vert=k$ and $\Vert A\bar{x}\Vert=k\Vert \lambda_1w_1\bar{v}_1 +\lambda_2w_2\bar{v}_2+...+\lambda_nw_n\bar{v}_n \Vert$
$\Vert A \Vert = \mathrm{sup}_{\bar{x}\neq 0}\frac{\Vert A\bar{x}\Vert}{\Vert \bar{x} \Vert}=\mathrm{sup}_{w_i} \Vert \lambda_1w_1\bar{v}_1 +\lambda_2w_2\bar{v}_2+...+\lambda_nw_n\bar{v}_n \Vert $
Now, notice that we can always select a unit vector $\bar{u}$ (with suitable $w_1,w_2,...,w_n$) such that
$$\Vert \lambda_1w_1\bar{v}_1 +\lambda_2w_2\bar{v}_2+...+\lambda_nw_n\bar{v}_n \Vert \ge \vert \lambda_{max} \vert$$
where $\lambda_{max}=max_i\{\lambda_i\}$
$\therefore \Vert A \Vert = \mathrm{sup}_{\bar{x}\neq 0}\frac{\Vert A\bar{x}\Vert}{\Vert \bar{x} \Vert}=\mathrm{sup}_{w_i} \Vert \lambda_1w_1\bar{v}_1 +\lambda_2w_2\bar{v}_2+...+\lambda_nw_n\bar{v}_n \Vert \ge\vert \lambda_{max} \vert \ge \vert \lambda_i \vert $