Polynomials of semi-simple linear operator is semi-simple

Since $T$ is semi-simple, its minimal polynomial is a product of distinct irreducible factors $q_1,\ldots,q_k$. Let $N_i=\ker\big(q_i(T)\big)$. Then $N_i$ is an invariant subspace of $p(T)$ and $V=N_1\oplus\cdots\oplus N_k$. Therefore, to prove that $p(T)$ is semi-simple, it suffices to prove that each $p(T|_{N_i})$ is semi-simple.

Suppose the contrary that $S=p(T|_{N_i})$ is not semi-simple. Then its minimal polynomial can be written as $f^2g$ for some non-constant polynomial $f$. Hence there exists some nonzero vector $x\in N_i$ such that $f(S)g(S)x\ne0=f(S)^2g(S)x$. If we put $y=f(S)g(S)x$, this means $y\ne0$ and $(f\circ p)(T|_{N_i})y=f(S)y=0$. However, as the minimal polynomial $q_i$ of $T|_{N_i}$ is irreducible, it is the minimal polynomial of every nonzero vector in $N_i$ w.r.t. $T|_{N_i}$. Therefore $q_i$ divides $f\circ p$. But then $(f\circ p)(T|_{N_i})=0$ and we arrive at the contradiction that $$ 0\ne y=f(S)g(S)x=(f\circ p)(T|_{N_i})g(S)x=0. $$ Hence $p(T|_{N_i})$ must be semi-simple.