TITLE: Prove a polynomial identity
QUESTION [5 upvotes]: Define a sequence of polynomials in the following way:
$P_m(t)=\frac {1} {m!}\cdot t\cdot (t-1)\cdot...\cdot (t-m+1) $.
(Where $P_0(t)=1$).
I'm trying to prove the following identity:
$\frac d {dt} P_{m+1}(t) = \sum_{k=0}^{m} \frac {(-1)^{m-k}} {m-k+1} \cdot P_k(t)$
Induction on $m$ doesn't seem to work here.
I've observed that if $f(x)=x^t$, then $P_m(t)=\frac {f^{(m)}(1)} {m!} $, but it didn't lead me anywhere.
Any ideas how to prove that?

REPLY [3 votes]: Generating functions to the rescue! Your observation and Taylor series show us that
$$
x^t = \sum_{m=0}^\infty P_m(t) (x-1)^m;
$$
this converges uniformly for $\frac12\le x\le\frac32$ say and for $t$ in any bounded range. Therefore we may differentiate in $t$ term-by-term:
\begin{align*}
\sum_{m=0}^\infty \frac d{dt}P_m(t) (x-1)^m &= \frac d{dt} x^t \\
&= x^t \log x \\
&= \bigg( \sum_{m=0}^\infty P_m(t) (x-1)^m \bigg) \bigg( \sum_{m=1}^\infty \frac{(-1)^{m-1}}{m} (x-1)^m \bigg).
\end{align*}
Your identity then follows by comparing the coefficients of $(x-1)^{m+1}$ on both sides.