Differentation Operator having trouble completing the proof for this question

Let $D:\mathbb{R}[X] \to \mathbb{R}[X]$ be the differentiation operator $D(f(X))=f'(X) .$ Prove that $e^{tD}(f(X)) = f(X+t)$ for $t \in \mathbb{R}$ 

Im having trouble making sense of the question. At first i tried Taylor's theorem to try and make sense of, and equate the two sides of the equation. This approach hasn't really worked. But could i go a more algebraic route using the fact that there exists a matrix D that represents this operator, and we know that $X^n$ spans D.
 A: You are right in the sense that it is the Taylor Theorem. To be more accurate and to prove it :
For every polynomial $P(X)$ of degree n :
$$P(X+t)=\sum_{k=0}^n \frac{P^{(k)}(X)}{k!} t^k=\left(\sum_{k=0}^n \frac{D^k}{k!} t^k\right) P(X)$$
This precisely means that the polyomial in X $P(X+t)$ is equal to the polynomial (in X still !) on the right of the previous expression.
By definition $e^{tD}=\sum_{k=0}^\infty \frac{D^k}{k!} t^k$. Because $D^{n+1}P=0$ for P of degree less than n, the exponential is reduced to the previous expression which shows that for any $P$ of degree n
$$P(X+t)=e^{tD}P(X)$$
(Once again, be careful about the fact that these two quantities should be considered as polynomial in X)
A: Noting (thanks go to @copper.hat) that the Taylor series of $e^{tD}$ is a sum on polynomials (as mentioned in the comments above), for every polynomial $p(X)=\sum_{k=0}^m a_kX^k$ one has
$$e^{tD}p(X) = \sum_{n=0}^\infty \frac{1}{n!}t^nD^n\left(\sum_{k=0}^m a_kX^k\right) = \sum_{n=0}^m \frac{1}{n!}t^nD^n\left(\sum_{k=0}^m a_kX^k\right) = \sum_{k=0}^m a_k\left(\sum_{n=0}^m \frac{1}{n!}t^nD^nX^k\right) = \sum_{k=0}^m a_k\left(\sum_{n=0}^k \frac{1}{n!}t^nD^nX^k\right)$$
Can you complete the proof?
