TITLE: Exponential generating function of the falling factorial
QUESTION [7 upvotes]: Let $\alpha$ be a real number. Define the sequence $(a_n)_n$ by $a_0=1$ and $a_n=\alpha(\alpha-1)\cdots(\alpha - (n-1))$ for $n\geq 1$. Find the exponential generating function of this sequence.
We have that $a_n=(\alpha-(n-1))a_{n-1}$ for $n\geq1$, so
\begin{align*}
A(x)&=\sum_{n\geq 0}a_n\frac{x^n}{n!}=a_0+\sum_{n\geq 1}a_n\frac{x^n}{n!}\\
&=1+\sum_{n\geq 1}(\alpha+n-1)a_{n-1}\frac{x^n}{n!}\\
&=1+\sum_{n\geq 0}(n+\alpha)a_{n}\frac{x^{n+1}}{(n+1)!}=1+\alpha\int_0^xA(t)dt+\sum_{n\geq 0}n\frac{x^{n+1}}{(n+1)!}\end{align*}
I'm stuck here. I tried to write the last sum as an integral and then solve a differential equation for $A(x)$, but it didn't work.
Should I search for another recurrence relation that $a_n$ satisfies?

REPLY [5 votes]: This is just the binomial series
$$\sum_{n\ge0}\alpha^{\underline n}\frac{x^n}{n!}=
\sum_{n\ge0}\frac{\alpha^{\underline n}}{n!}x^n=
\sum_{n\ge0}\binom{\alpha}nx^n=(1+x)^\alpha.
$$