Let $n\in \mathbb N$ and let $N$ be an $n\times n$ matrix over the field $F$ such that $N^n=0$ but $N^{n-1}\neq 0$. Prove that there is no $n\times n$ matrix $A$ such that $A^2=N$.
I can understand that $N$ is nilpotent, and if there is such an $A$, then $A^{2n}=0$ which forces $A$ to be nilpotent as well. But I can't see any obvious way to use it to prove that such an $A$ doesn't exist. Also, I can't see how to relate the order of a matrix with the exponent.
It is assumed that $n \neq 1$. $A^{2n}=0$ and this implies that $A^{n}=0$ (because $A$ is $n \times n$). Hence, $A^{k}=0$ for all $k \geq n$. But $2n-2 \geq n$ so $A^{2n-2}=0$. So $N^{n-1}=0$, a contradiction.
[To show tat $A^{n}=0$ observe that $0$ is the only eigne value. Use the fact that $p(A)=0$ where $p$ is the characteristic polynomial (Cayley Hamilton Theorem:
https://en.m.wikipedia.org/wiki/Cayley%E2%80%93Hamilton_theorem )].