An $n\times n$ nilpotent matrix is nilpotent of degree at most $n$ The following is a common linear algebra fact that has been asked for specific choices of $n\in\mathbb N$ in the past (see 1, 2, 3, 4, 5 among others.)

$\DeclareMathOperator{\kk}{\mathbb k}$Suppose that $A$ is a $n\times n$ matrix with coefficients in a field $\kk$ and that it is nilpotent, that is, there exists a non-negative integer $k$ such that $A^k=0$. Then in fact $A^n=0$.

Disclaimer: For the sake of having an authoritative post to reference and use to close future (abstract) duplicates this post was edited accordingly. This was done with particular interest in the highest voted (but not accepted) elementary proof which may be missed by students, who may also be quick to appeal to the Jordan normal form (or the minimal polynomial say), which require unnecessary machinery and obscure the proof.
For example, this post implies immediately that the characteristic polynomial of $A$ is $X^n$ over any field and hence that the minimal polynomial is $X^r$ for some $1\leqslant r\leqslant n$ which is mentioned in the accepted answer.
 A: If a matrix is nilpotent, its minimal polynomial is $x^r$  for some $r$. Now this minimal polynomial is a divisor of its characteristic polynomial, which has degree $\dim A$.
A: Here's a longer answer. Say $A^k=0$. If $k\leq n$ we're done, so let's suppose $k>n$. Consider the nullspaces of successive powers of A. We have
$$\ker(A)\subseteq \ker(A^2) \subseteq \cdots \subseteq \ker(A^n)$$
Since $A$ is nilpotent, $\det(A)=0$ so $\ker(A)\neq 0$. If $A^n \neq 0$ then $\dim\ker(A^n) < n$, so that
$$0 \subseteq \ker(A) \subseteq \ker(A^2) \subseteq \cdots \subseteq \ker(A^n) \subseteq \mathbb k^n.$$
These are $n$ subspaces $V_i = \ker(A^i)$ and $n-1$ possible values for their dimensions. The pigeonhole principle implies there exists $1\leqslant m < n$ such that $\ker(A^n) = \ker(A^m)$.
Now for all $v \in \mathbb{R}^n$ and $t \in \mathbb{N}$ we have that , $A^t v=0$ implies
$A^{t - (m+1)}v \in \ker(A^{m+1})= \ker(A^m)$ and so $A^{t-1}v=0$. Repeat the argument $k-n$ times with $A^kv=0$ to conclude that $A^nv=0$ for any $v$, which is what you wanted.
A: Somehow similar to the to the first answer. Take $f(t)$ the characteristic polynomial of $A$, some $n \times n$ matrix . In some algebraic closure of your original field, you find the eigenvalues. Then since every eigenvalue of the your matrix must be zero (since A is nilpotent),$f(t)=(-1)^n t^n$, then by Cayley-Hamilton $f(A)=0$, so the degree is at most $n$.
