If $\lambda$ is an eigenvalue of $A^k$, prove that at least one of the $k^{th}$ roots of $\lambda$ is an eigenvalue of $A$.
I know it is easy to prove when $A$ can be presented by a matrix, but $A$ is an operator on a complex inner product space, which may be infinite dimensional.
Let $\mu_1,...,\mu_k$ be the roots of order $k$ of $\lambda$.
Clearly $\mu_1,...,\mu_k$ are solutions for the polynomial $(x^k-\lambda)$. In particular we have $x^k-\lambda = \prod_{l=1}^k (x-\mu_l)$.
We conclude that $(A^k-\lambda\cdot I) = \prod_{l=1}^{k} (A-\mu^l\cdot I)$.
Can you finish now? (If $\ker(S)=\ker(T)=\{0\}$ what can you say about $\ker(ST)$?)