Inverse operator of $I-A$ Let $H$ be an Hilbert space, $A:H\to H$ be a bounded linear operator such that
$$
\|A^{n_0}\|< 1\qquad\text{for some}\quad\; n_0\in\mathbb{N}.
$$
I have to show that $I-A$ is invertible.
My idea is to consider 
$$
S:=\sum_{n=0}^\infty A^n,
$$
where $A^n$ is defined by induction as usual.
Then
$$
(I-A)S=\lim_{n\to\infty} I-A^{n+1}.
$$
We have just to show 
$$
\lim_{n\to\infty}A^{n+1}=0.
$$
Any ideas?

Hint:
\begin{align}
S = \sum_{m=0}^{\infty} \sum_{\ell = 0}^{n_{0} -1} A^{mn_{0}+\ell}
\end{align}
