TITLE: Show $e^{D}(f(x)) = f(x+1)$ where $D$ is the derivative operator
QUESTION [8 upvotes]: I would appreciate help showing $e^{D}(f(x)) = f(x+1)$
Where $D$ is the linear operator $D: \mathbb{C}[x] \rightarrow \mathbb{C}[x]$ where (in the context where this statement arose) $x \in \mathbb{N}$; 
$f(x) \mapsto \frac{d}{dx} f(x)$
By the Taylor series expansion $e^{D} = \sum_{n=0}^{\infty} \frac{D^n}{n!}$
$(1)$ Then $e^{D} (f(x)) = f(x) + f'(x) + \frac{f''(x)}{2!} +\dots$
I would appreciate help showing that the above line $(1)$ is equal to $f(x +1)$
This is what I have tried, but it feels forced:
for a Taylor series representation of $f(x +1)$ near $x$ I could write
$f(x+1) = f(x) + f'(x)(x+1 - x) + \frac{f''(x)}{2!}(x+1 -x)^2 \dots$
This then is equal to the RHS of line $(1)$.
Could this line of thinking be correct? Thanks very much.

REPLY [3 votes]: Use the binomial theorem.
$$
e^D x^n = \sum_{k=0}^\infty \frac{D^k}{k!} x^n.
$$
All of the terms after $k=n$ are $0$, i.e. $D^k x^n=0$ if $k>n$, so this is
\begin{align}
\sum_{k=0}^n \frac{D^k}{k!} x^n & = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!} x^{n-k} \\[6pt]
& = \sum_{k=0}^n \binom nk x^{n-k} \\[6pt]
& = (x+1)^n.\quad\longleftarrow\text{By the binomial theorem.}
\end{align}