Let $A$ be a real-valued, square matrix and define its 2-norm as:
$$||A||_2 = \sqrt{\max_i\lambda_i(AA^T)}$$
where $\lambda_i(AA^T)$ denotes the $i^{th}$ eigenvalue of the product $AA^T$.
Now presume I have a product of $k$ matrices:
$$\phi_k = A_1A_2 \cdots A_k, k \in [0,1,...,k]$$
Since the 2-norm is submultiplicative, I know that, for finite $k$:
$$||\phi_k||_2 \le ||A_1||_2\times||A_2||_2\times \cdots \times ||A_k||_2$$
Let us now further assume that $||A_k||_2 \lt 1 \forall k$.
Is this statement true:
$$||\lim_{k \rightarrow \infty} \phi_k||_2 = 0$$
I suspect that it is, but that's only because I'm extrapolating the condition for the norm of $\phi_k$ when $k$ is finite. I know it is true if all of the $A_k$'s were the same, since I'd just do an eigendecomposition (remember my matrices are both real and square). However, it's the case where the $A_k$'s are different that causes me alarm.
If I have understood your question correctly, then your statement fails to hold for $1 \times 1$ matrices, let alone the $n \times n$ case. In particular, if we take $$ a_k = e^{-1/2^k}, \quad k = 0,1,\dots $$ Then we have $\lim_{k \to \infty}\|\phi_k\| = \lim_{k \to \infty}\prod_{j=1}^k a_j = e^{-2} \neq 0$.