Bound on the norm of a matrix power Suppose we have the square matrix $A$ and we know that its spectral radius $\rho(A)$ is less than $1$, therefore matrix $A$ is stable. How can we prove that $\exists \gamma \in(0,1)$ and $\exists M >0$ such that 
$$\|A^k\|\leq M\gamma^k, \:\:\:\: \forall k\geq0$$
What I tried so far is $\|A^k\|=\|A\dots A\|\leq\|A\|\dots \|A\| =\|A\|^k$ so taking $\gamma=\|A\|$ I should be close to the above inequality, but I am not sure it is correct.

Let $\lVert \cdot \rVert$ be a matrix norm on $\mathbb{C}^{n \times n}$. I assume that we know Gelfand's formula:
$$
\rho(A) = \lim\limits_{k\to\infty}\lVert A^k\rVert^{1/k}.
$$
We want to prove that for any $\gamma > \rho(A)$ there exists $M = M(\gamma) \ge 1$ such that
$$
\lVert A^k \rVert \le M \gamma^k, \quad k \in \mathbb{N}.
$$
It follows from Gelfand's formula that there exists $k_0$ such that for any $k = k_0 +1, k_0 + 2, \ldots,$ there holds
$$
\lVert A^k \rVert < \gamma^k.
$$
It suffices now to take
$$
M := \max(\{1\} \cup \{\lVert A^k \rVert/\gamma^k : k = 1, \ldots, k_0 \}).
$$
When $\rho(A) < 1$ we can take $\gamma \in (\rho(A), 1)$.
