degree of nilpotent matrix

I need to prove or disprove that A nilpotent matrix's degree is less than or equal to its dimension. I tried to make a counterexample but I found nothing and I think that this claim is true but I don't know the start point!

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

$N(A)\subseteq N(A^2) \subseteq ... \subseteq N(A^n)$

Since $A$ is nilpotent, $det(A)=0$ so $nullity(A)>0$. If $A^n \neq 0$, $nullity(A^n)<n$. Thus

$1 \leq nullity(A) \leq nullity(A^2) \space... \space \leq nullity(A^n) \leq n-1$

Here are $n$ matrices with $n-1$ possible values for nullity. Pigeonhole principle implies $\exists m<n$ $nullity(A^m) = nullity(A^{m+1})$. Now $N(A^m) \subseteq N(A^{m+1})$ implies $N(A^m) = N(A^{m+1})$.

Now $\forall v \in \mathbb{R}^n$ and $l \in \mathbb{N}$, $A^lv=0$ implies $A^{l-(m+1)}v \in N(A^{m+1})=N(A^m)$ and so $A^{l-1}v=0$. Repeat the argument $k-n$ times with $A^kv=0$ to conclude that $A^nv=0$ for any $v$.

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 $\dim A$.