I have asked a long time ago a question about matrix exponential shifting. I considered the following two operators $D$ and $S$ in the vector space $\mathcal{P}_{n}$ of polynomials of degree $\leq n$ :
$$ D f(t)=\frac{d f}{d t}(t) \quad \text { and } \quad S f(t)=f(t+1),$$
for each $f\in\mathcal{P}_{n}$.
We established the identity $\exp(D)=S$. Now I would like to determine the matrices of operators $D$ and $S$ in the basis $\{1, t, t^{2}, \ldots, t^{n}\}$ of $\mathcal{P}_{n}$. This is what I have so far:
$f\in\mathcal{P}_{n}$ has the form
$$ \sum_{k=0}^n a_kt^k=a_0+a_1t+a_2t^2+...+a_{n-1}t^{n-1}+a_nt^n. $$
The operator $D$ maps this to $$ a_1+2a_2t+...+(n-1)a_{n-1}t^{n-2}+na_nt^{n-1}= \sum_{k=1}^n ka_kt^{k-1} $$ The new coefficients for the $j$-th basis vector are in the $j$-th column, so the mapping matrix has the form $D=\left(\begin{array}{ccccccc}0 & 1 & 0 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 2 & 0 & \cdots & 0 & 0 \\ 0 & 0 & 0 & 3 & \ddots & \vdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & 0 & \vdots \\ 0 & 0 & 0 & \cdots & 0 & n-2 & 0 \\ 0 & 0 & 0 & 0 & \cdots & 0 & n-1 \\ 0 & 0 & 0 & 0 & \cdots & 0 & 0\end{array}\right)$.
My problem is now to find the Matrix of $S$. I thought I can use the identity $\exp(D)=S$ and the good thing is, that
$$D^n=0_n$$
and I can use that. But it I dont know how to prove it and I wanted to ask if there is
a) a better way to find the matrix of $S$
b) an easy way to show my assumption above, maybe with this summation style (I do not really get how though).
Thanks in advance for any type of hint or help.
Hint:
$$\mathbf S\left[X^k\right] = (1+X)^k = \sum_{j=0}^k \binom{k}{j}X^j$$