I'm trying to figure out the very last sentence of the proof of this theorem, and I don't understand why what they're saying is true.
Theorem: Let $V$ be an $F$ vector space with $\dim_{F} V=n$ and $T:V \longrightarrow V$ linear with characteristic polynomial $f_T$ where $\lambda$ is an eigenvalue of $T$. Then if $f_T$ splits in $F[X]$ and the algebraic multiplicity of $\lambda$ is $m$, then:
$\dim_{F} K_{\lambda}(T) \leq m$ where $K_{\lambda}(T)$ is the generalized eigenspace of $T$ for $\lambda$.
$N((T-\lambda I)^m) = K_{\lambda}(T)$.
I follow everything up until showing $K_{\lambda}(T) \subset N((T-\lambda I)^m)$. The proof given just says 'it follows immediately by the choice of $m$.' But I don't see how it follows at all. If $x \in K_{\lambda}(T)$, then there exists some $p \in \mathbb{N}$ such that $(T-\lambda I)^p(x) = 0$, but I don't see why the dimension of $K_{\lambda}(T)$ being bounded above by $m$ proves this containment.