Eigenvector of polynomial 
Suppose that $T: V \rightarrow V$ is an endomorphism of the linear space V (about $\mathbb{K}$) and that $p(X)$ is a polynomial with coefficients in $\mathbb{K}$. Show that if $x$ is an eigenvector of $T$ than it is also an eigenvector of $p(T)$.

My attempt:
So if $x$ is an eigenvector of $T$ that means that $T(x) = \lambda x$ ($\lambda$ being the eigenvalue associated to $x$).
Ok so my next step is the one I feel is not correct 
$p (T(x)) = p (\lambda x)$ so $\lambda$ is an eigenvalue of the polynomial.
I don't feel this is a correct assumption, that you can't immediately conclude this, can  we?

Here is an outline.
Step 1: Prove by induction that if $Tx=\lambda x$, then $T^n(x)=\lambda^n x$.
Step 2: Show that if $Tx=\lambda x$ and $Sx=\mu(x)$, then $(T+S)(x)=(\lambda+\mu)x$.
Step 3: Combine the first two steps to show that if $Tx=\lambda x$, then $P(T)(x)=P(\lambda)x.$
