Let $K$ be a field of characteristic $0$, $f \in K[t]$ a polynomial over $K$ and $J \in M_{n,n}(K)$ a Jordan block to an eigenvalue $\lambda \in K$, meaning that $J$ has the shape:
$$J = \pmatrix{\lambda & 0 & ... & 0 \cr 1 & \lambda & ... & 0 \cr 0 & 1 & \lambda & ... \cr & ...&...& \cr 0 & ... & 1 & \lambda}$$
I now want to show that
$$f(J) = \pmatrix{f(\lambda) & 0 & ... \cr f'(\lambda) & f(\lambda) & 0 & ... \cr \frac{1}{2} f''(\lambda) & f'(\lambda) & f(\lambda) & 0 &... \cr ... \cr \frac{1}{(n-1)!}f^{(n-1)}(\lambda) & ... & \frac{1}{2}f''(\lambda) & f'(\lambda) & f(\lambda)}$$
Where $f', f'', ..., f^{(n-1)}$ denote the derivatives of $f$.
My first thought was to write $J = \lambda E_n + N$, with $N$ being nilpotent; and maybe work out the potencies of $J$ thereafter. But I don't know how to continue from there to get to the required shape of $f(J)$.
We have $$ (\lambda I + N)^k = \sum_{j=0}^k \binom kj \lambda^{k-j}N^j $$ Now, taking $f(t) = \sum_{k=0}^m a_k t^k$, we have $$ \sum_{k=0}^m a_k(\lambda I + N)^k = \sum_{k=0}^m a_k\sum_{j=0}^k \binom kj \lambda^{k-j}N^j =\\ \sum_{j=0}^k \left(\sum_{k=0}^m a_k \binom kj \lambda^{k-j}\right) N^j =\\ \sum_{j=0}^k \left(\frac 1{j!}f^{(j)}(\lambda)\right) N^j $$