Determine the null space of a linear map Let $P_k(x)$ denote the space of polynomials of at most degree $k$. Let $D$ denote differentiation with respect to $x$. Regard the differential operator $L: P_k\rightarrow P_k$ such that $L=\frac{1}{n!}D^n+\frac{1}{(n-1)!}D^{n-1}+...+D+I$ . If $k\leq n$, find the dimension of the kernel of $L-T$ where $T:P_k\rightarrow P_k$ is given by $T(p(x))=p(x+1)$. 
To minimize the amount of calculation, I start with finding the matrix representation of $D$ w.r.t $\{1,x,x^2,...,\}$ basis, which is a matrix with $1,2,3,..,n$ on the super diagonal and 0 everywhere else. Then should I find $D^k$ for each $k$? The computation seems to be insane. Are there any easier way? Any shortcuts?
 A: Hint: Taylor's formula tells us that when $p(x)$ is a polynomial:
$$p(x+1)=p(x)+p'(x)+\frac{1}{2}p''(x)+\cdots+\frac{1}{n!}p^{(n)}(x)+\cdots$$
since the remainder (Lagrange, etc.) will eventually become zero. Write this formula in the form of operators on $p$.
A: Hint Since $D: P_k \to P_k$ is nilpotent of order $k + 1$, that is, $D^l = 0$ for $l \geq k + 1$, the series defining its exponential truncates at that order. For $n \geq k + 1$ we have:
$$
L = I + D + \frac{1}{2} D + \cdots + \frac{1}{n!} D^n
  = \sum_{i = 0}^{\infty} \frac{1}{i!} D^i
  =: \exp D.
$$
For the matrix representation $[D]$ of $D$ w.r.t. the given basis $\{1, x, \ldots, x^k\}$, it's tractable to work out manually a change-of-basis matrix that puts $D$ in Jordan Normal Form, after which it is easy to exponentiate.

Additional hint The matrix representation for $L$ with respect to the given basis has $(i, j)$ entry $j \choose i$ for $j \geq i$ and $0$ for $j < i$, that is, it is upper triangular with nonzero entries given by (the transpose of) the first $k + 1$ rows of Pascal's Triangle.

