Why are eigenvalues of nilpotent matrices equal to zero?

If $v$ is a non-zero eigen vector corresponding to an eigenvalues $\lambda$ we have, by definition, $Av=\lambda v$. Then $A^2v= A( Av)= A(\lambda v)= \lambda^2v$. It easily follows that $\lambda^n$ is an eigenvalue for $A^n$ but the latter is the zero matrix, for which all eigenvalues are zero, hence $\lambda=0$.

Suppose $\lambda$ is an eigenvalue of the nilpotent matrix $A,$ and $u$ its associated eigenvector. Then $$Au = \lambda u, u \neq 0$$ multiplying by $A$ on the right shows $$A^2u = \lambda Au = \lambda^2 u$$ and by induction $$A^n u = \lambda^n u$$

If $A$ is nilpotent, then $A^k = 0$ for some $k>0.$ that implies $\lambda^k = 0\to \lambda = 0.$

Alternatively, do the contrapositive. If $A$ has a non-zero eigenvalue, $\lambda,$ then $A^{k} \neq 0 $ for all $k.$

Proof: there exists $v \neq 0,$ such that $$ Av = \lambda v, $$ so $$ A^{k} v = \lambda^{k} v \neq 0. $$ Done.