Why does the $n$-th power of a Jordan matrix involve the binomial coefficient?

Let $N$ denote the nilpotent matrix whose superdiagonal contains ones and all other entries are zero. Then $N^k=0$. Hence, by the binomial theorem: $$J_k(\lambda)^n=(\lambda I+N)^n=\sum_{r=0}^\color{red}{n} \binom{n}{r}\lambda^{n-r} N^r=\sum_{r=0}^\color{red}{\min(n,k-1)} \binom{n}{r}\lambda^{n-r} N^r.$$

user1551 already gave the "simple explanation", so let me just mention the general result for evaluating matrix functions when the argument is a Jordan block. Given the $n\times n$ Jordan block with eigenvalue $\lambda$,

$$\mathbf J=\begin{pmatrix}\lambda&1&&\\&\lambda&\ddots&\\&&\ddots&1\\&&&\lambda\end{pmatrix}$$

the matrix $f(\mathbf J)$ looks like this:

$$f(\mathbf J)=\begin{pmatrix}f(\lambda)&f^\prime(\lambda)&\cdots&\frac{f^{(n-1)}(\lambda)}{(n-1)!}\\&f(\lambda)&\ddots&\vdots\\&&\ddots&f^\prime(\lambda)\\&&&f(\lambda)\end{pmatrix}$$

This general result is proven in a number of references; e.g. this.

For the power function $f(x)=x^k$, we have the general result (easily proven inductively):

$$\frac1{j!}\frac{\mathrm d^j}{\mathrm dx^j}x^k=\binom{k}{j}x^{k-j}$$

Making the necessary replacements gives the formula for the power of a Jordan block.