One of my exercises asks the following.
Let $D\colon\mathbb R[X]\to\mathbb R[X]$ (where $\mathbb R[x]$ is the space of polynomials with real coeffients) be the differential operator $D(f(X))=f'(X)$. Show that $$e^{tD}(f(X))=f(X+t)$$for any real number $t$.
EDIT: Using mgn's suggestion in the comments, we write $$e^{tD}(f(X)=\sum_{n=0}^\infty \frac{t^nD^n(f(X))}{n!}=f(X)+tf'(X)+\frac{t^2f''(X)}2 +\frac{t^3 f^{(3)}(X)}{6}+\cdots.$$Now, we fix $X$ and find the Taylor series of the function $f(X+t)$ about $t=0$: $$f(X+t)=f(X)+tf'(X)+\frac{t^2f''(X)}{2}+\cdots.$$ But this is precisely $e^{tD}(f(X))$ and we are done.
Can somebody please verify that this proof is rigorous and correct?
Thanks in advance.
Yes, you understood my comment correctly and the proof which you presented seems correct to me. For more details you could write something like: let $X$ be fixed then $g(t):=f(X+t)$ is still a polynomial thus we may apply Taylor's expansion about a point $t=0$. Thus $$f(X+t)=g(t) = g(0) + \sum_{n \geq 1} \frac{g^{(n)}(0)}{n!} t^n = \sum_{n \geq 0} \frac{f^{(n)}(X)}{n!} t^n =e^{tD}(f(X)),$$ where $f^{(0)} =f$.
Good job!