Edit: This answer is unfortunately incorrect. The error is highlighted below. The argument does work if we are allowed to transpose the matrices.
We start by proving a lemma.
Lemma. Let A be an n×n matrix and let N be the n×n matrix with ones on the secondary diagonal. If ANt and NtA are nilpotent for all t≥0 then A=0. Correct conclusion: A is upper triangular with zeroes on the diagonal. (The original conclusion is recovered if we are also allowed to multiply by powers of the transpose of N.)
Proof. Suppose for example that n=3, and write
A=⎛⎝⎜adgbehcfi⎞⎠⎟,N=⎛⎝⎜000100010⎞⎠⎟.
We start by calculating
AN2:
AN2=⎛⎝⎜000000adg⎞⎠⎟.
This matrix is in triangular form, and so if
AN2 is nilpotent then
g=0. Continue with
AN1:
AN1=⎛⎝⎜000adgbeh⎞⎠⎟=⎛⎝⎜000ad0beh⎞⎠⎟.
Again the matrix is in triangular form, and so if
AN1 is nilpotent then
d=h=0. Continuing,
AN0=⎛⎝⎜a00be0cfi⎞⎠⎟.
As before, we conclude that
a=e=i=0, and so
A is upper triangular with zeroes on the diagonal.
If we now consider N2A,N1A,N0A instead, then we conclude that A is lower triangular with zeroes on the diagonal. In fact, we don't get anything new from considering NtA. Therefore A=0. □
This is how the proof would go if the original version of the lemma were correct.
Now back to the problem at hand. Say that the matrices A1,…,Ak satisfy property P if for every infinite sequence i1,…∈[k], we have Ai1⋯Aim=0 for some m. If one of the matrices Ai is not nilpotent then property P clearly fails, so suppose that all the matrices are nilpotent. If all matrices commute then property P clearly holds, so suppose that A1A2≠A2A1. Change basis so that A1 is in Jordan normal form, and let the corresponding decomposition of the vector space be V1⊕⋯⊕Vt. Let Vi be a vector space on which A1A2≠A2A1; note that dimVi>1 since 0 commutes with everything. Restricted to Vi, A1=N and A2≠0. Therefore the lemma implies that for some t≥0, either A2At1 or At1A2 is not nilpotent, and therefore property P clearly fails.
Summarizing, property P holds iff all matrices are nilpotent and all of them commute.