Prove that the only eigenvalue of a nilpotent operator is 0?

$\phi$ is nilpotent, so $\phi^n = 0$ for some $n$. Now let $v$ be an eigenvector: $\phi v = \lambda v$ for some scalar $\lambda$. Now we get $$ 0 = \phi^n v = \lambda^n v ~\Rightarrow~ \lambda=0 ~. $$

Note that this works in the infinite-dimensional case as well; there is no need to relate $\phi$ to a matrix.

Edit: As suggested in the comments, we can also show that $0$ is always an eigenvalue; in other words, $\phi$ always has at least one eigenvector. Take any $v \neq 0$; we know that $\phi^n v = 0$, so let $k$ be the largest integer such that $\phi^k v \neq 0$. Then $\phi(\phi^k v) = 0$, so $\phi^k v$ is an eigenvector of $\phi$, with eigenvalue $0$.

Suppose that $\phi$ has another eigenvalue $\lambda \ne 0$ so that $\phi(x)=\lambda x $ ($x$ is an eigenvector corresponding to $\lambda$)

Then, $ \phi^n (x)= \phi^{(n-1)}(\phi(x))=\phi^{(n-1)}(\lambda x)=\lambda \phi^{n-1}(x)=\cdots=\lambda ^{n}x\ne 0$.

We have a contradiction, so $\phi$ can't have another eigenvalue except $0$.