TITLE: Proof that the binomial transform is involution
QUESTION [5 upvotes]: Let $\{x_n\}$, $n=0,1,2,\ldots$, be a sequence, and let $\{y_n\}$, $n=0,1,2,\ldots$, be its binomial transform, that is,
$$
y_n=\sum_{k=0}^{n} (-1)^k {n\choose k} x_k.
$$
I need to prove that the binomial transform is an involution, that is,
$$
x_n=\sum_{k=0}^{n} (-1)^k {n\choose k} y_k.
$$
I tried to use the combinatorial Vandermonde's identity but I failed. Please help me to prove.

REPLY [2 votes]: It is also convenient to show the involution property with the help of exponential generating functions.

Let
  \begin{align*}
A(t)=\sum_{n=0}^\infty x_{n}\frac{t^n}{n!} \qquad\text{and}\qquad B(t)=\sum_{n=0}^\infty y_{n}\frac{t^n}{n!}
\end{align*}
  the exponential generating functions of  $\{x_n\}$ and $\{y_n\}$.

Since
\begin{align*}
A(-t)e^t&=\left(\sum_{k=0}^\infty x_{k}\frac{(-t)^k}{k!}\right)\left(\sum_{l=0}^\infty \frac{t^l}{l!}\right)\\
&=\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{(-1)^kx_k}{k!}\cdot\frac{1}{l!}\right)t^n\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}(-1)^kx_k\right)\frac{t^n}{n!}\\
&=\sum_{n=0}^\infty y_n\frac{t^n}{n!}\\
&=B(t)
\end{align*}

we obtain the following correspondence
  \begin{align*}
B(t)=A(-t)e^t\qquad\longleftrightarrow\qquad y_n=\sum_{k=0}^n\binom{n}{k}(-1)^k x_{k}\tag{1}
\end{align*}

Multiplication with $e^{-t}$ and substitution $t$ with $-t$ gives
\begin{align*}
A(t)=B(-t)e^{t}
\end{align*}

According to (1) we conclude due to symmetry
\begin{align*}
A(t)=B(-t)e^t\qquad\longleftrightarrow\qquad x_n=\sum_{k=0}^n \binom{n}{k}(-1)^ky_{k}
\end{align*}
  and the claim follows since sequence and its binomial transform have the same generating function.