I've searched a lot for a **simple explanation** of this. Given a Jordan block $J_k(\lambda)$, its $n$-th power is:

$$ J_k(\lambda)^n = \begin{bmatrix} \lambda^n & \binom{n}{1}\lambda^{n-1} & \binom{n}{2}\lambda^{n-2} & \cdots & \cdots & \binom{n}{k-1}\lambda^{n-k+1} \\ & \lambda^n & \binom{n}{1}\lambda^{n-1} & \cdots & \cdots & \binom{n}{k-2}\lambda^{n-k+2} \\ & & \ddots & \ddots & \vdots & \vdots\\ & & & \ddots & \ddots & \vdots\\ & & & & \lambda^n & \binom{n}{1}\lambda^{n-1}\\ & & & & & \lambda^n \end{bmatrix}$$

Why does the $n$th power involve the binomial coefficient?