This is the problem 35.3 from Halmos's 'Finite-Dimensional Vector Spaces'.
Let $\mathscr{P}_n$ be the set of all polynomials of degree less or equal to $n-1$, having complex coefficients (together with the identically zero polynomial, $\mathscr{P}_n$ forms a vector space).
Let $A$ be a linear transformation on $\mathscr{P}_n$, such that $Ax(t) = x(t+1)$ for every $x \in \mathscr{P}_n$. Prove that if $D$ is the differential operator, then \begin{gather} \tag{$\star$} 1+\frac{D}{1!}+\frac{D^2}{2!}+\cdots+\frac{D^{n-1}}{(n-1)!}=A \label{eq} \end{gather}
Proof
The proof follows by induction. The base case for $n=1$ is straightforward to verify. Suppose that \eqref{eq} holds for $n$. Then, for $n+1$, observe that any polynomial $x$ in $\mathscr{P}_{n+1}$ can be written as $x=y+\alpha_n t^n$, with $y \in \mathscr{P}_n$. Then, \begin{align} Ax=A\big(y+\alpha_n t^n\big)&=y(t+1)+A(\alpha_nt^n)\\ &=y(t+1)+\alpha_n\big(t^n+\frac{n!}{1!(n-1)!}t^{n-1}+\frac{n!}{2!(n-2)!}t^{n-2}+\cdots+1\big)\\ &=y(t+1)+\alpha_n(t+1)^n\\ &=x(t+1) \end{align}
I) Is the proof correct?
II) I would be interested in an alternative -more intuitive- proof for the above problem. For example, the polynomial expansion of the operator $A$ resembles the one of the exponential function. Is there anything in common there? What if we had a more general translation, like $x(t+c)$? Again, I can see what's happening when I carry out the computations, but I wonder if there is a more intuitive proof, unveiling something more interesting.
Thanks in advance!
It is perhaps slightly more intuitive to use the fact that these are linear operators and so we only need prove identities for individual powers of $t$.
Consider the infinite series for $e^D$. Acting on $t^k$ it is immediate that this produces $(t+1)^k$.
Re. the question about $x(t+c)$, the same argument proves this transform is $e^{cD}$.
Hope this helps.