Hilbert space $\mathbb{L^2}$. So my thoughts:
$n=0$ - true. Let $\hat\alpha = (\hat{L^+})^n (\hat{L})^n $
For $n+1$: $\ \hat{L^+} \hat\alpha \hat{L} \phi = \lambda \phi = \hat\alpha \phi$ since its just unitary. But for $n=0$ eigenvalues are just $1$, no? So following my proof for $(\hat{L^+})^n (\hat{L})^n$ eigenvalues are $1$ too.
upd: which are false as I can see from now.
By induction. Base case $n=0$ is trival. Set $M = (L^*)^{n-1} L^{n-1}$. $M \ge 0$ by the induction hypothesis. We have $(L^*)^{n} L^{n} = L^* M L$, and $$ \langle L^* MLx, x\rangle = \langle M(Lx), Lx\rangle \ge 0, $$ which finishes the inductive step.