Answer
See below
Work Step by Step
Assume that $A_1, A_2,..., A_k$ are $n × n$ matrices
We obtain $A_n.v=\lambda_1 .v$ with $i=1,2,...k$
then $(A_1.A_2...A_k)v$$=(A_1.A_2...A_{k-1})(A_k.v)\\
=(\lambda_1.\lambda_2...\lambda_k)v$
Hence $v$ is also an eigenvector of the matrix $(A_1.A_2...A_k)$
The eigenvector: $=\lambda_1 \lambda_2...\lambda_k$
