A matrix is a product of nilpotent matrices iff its not invertible I just completed a homework problem which proves the following result: a matrix (with coefficients in some field) is a product of nilpotent matrices iff its not invertible.
The proof was broken into several parts and was quite involved. I'm wondering if there's a very simple way to demonstrate this result (just the implication non-invertible then its a product of nilpotent matrices , as the other implication is trivial).

Following is a proof for fields that allow a Jordan-type canonical form. Write $A=P^{-1}TP$ with $T$ in Jordan form (but this can be relaxed somewhat). $A$ is not invertible if, and only if, at least one of the diagonal elements is zero. Taking it to be the last eigenvalue for simplicity, note the following example decomposition into two nilpotent matrices: $$\pmatrix{1&2&3&4&0\\0&5&6&7&0\\0&0&8&9&0\\0&0&0&a&0\\0&0&0&0&0}=\pmatrix{0&1&2&3&4\\0&0&5&6&7\\0&0&0&8&9\\0&0&0&0&a\\0&0&0&0&0}\pmatrix{0&0&0&0&0\\1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0}$$
Thus $T=M_1M_2$ nilpotent, so $A=(P^{-1}M_1P)(P^{-1}M_2P)=N_1N_2$.
