# Expand generating function $\frac{\exp{\frac{z}{1-z}}}{1-z}$

I know that the coefficients $[z^n]$ of the exponential generating function $\frac{\exp{\frac{z}{1-z}}}{1-z}$ are $\sum_{i=0}^{n}i!\binom{n}{i}^2$ but have trouble in proving it. I have done the following:$$\exp{\frac{z}{1-z}}=\sum_{k=0}^\infty \frac{\big(\frac{z}{1-z}\big)^k}{k!}=\sum_{k=0}^{\infty}\frac{\sum_{i\geq k}\binom{i-1}{k-1}z^i}{k!}=\sum_{n=0}^{\infty}\bigg(\sum_{k=1}^n\frac{\binom{n-1}{k-1}}{k!}\bigg)z^n$$ and thus $$[z^n]\frac{1}{1-z}\exp{\frac{z}{1-z}}=\sum_{j=0}^{n}\sum_{k=1}^{j}\frac{\binom{j-1}{k-1}}{k!}$$ I have tried several things to turn the above sum into the wanted form but did not succeed. Is there anyone who can help?

• Here's the OEIS sequence with lots of references. – joriki Nov 10 '15 at 20:17
• You used $[z^n]$ for the coefficients of the exponential generating function at the top but for those of the ordinary generating function at the bottom, so the coefficients differ by a factor of $n!$. (Conventionally that notation is used for the coefficients of the ordinary generating function.) Note that what remains of the expression you're trying to obtain if you divide it by $n!$ is $\sum_{i=0}^n\frac1{i!}\binom ni$ (no square). – joriki Nov 10 '15 at 20:58

As I wrote in a comment, the expression at the top should be divided by $n!$ and then becomes $\sum_{i=0}^n\frac1{i!}\binom ni$. To get this from the expression at the bottom, exchange the order of summation and apply the hockey stick identity.