Let $A_{k, n}$ denote the number of idempotent functions on a set of size $n$ with $k$ fixed points. We have the formula $A_{k, n}=\binom{n}{k} k^{n-k}$ for $1 \leq k \leq n$ because there are $\binom{n}{k}$ ways to choose the fixed points and all $n-k$ remaining elements must map to fixed points, which can happen in $k^{n-k}$ ways. Hence $\sum_{n=1}^{\infty} \frac{I_{n}}{n!} =\sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{A_{k, n}}{n!} =\sum_{n=1}^{\infty} \sum_{k=1}^{n} \frac{k^{n-k}}{k!(n-k)!} =\sum_{k=1}^{\infty} \frac{1}{k!} \sum_{n=k}^{\infty} \frac{k^{n-k}}{(n-k)!} =\sum_{k=1}^{\infty} \frac{1}{k!} \sum_{n=0}^{\infty} \frac{k^{n}}{n!} =\sum_{k=1}^{\infty} \frac{1}{k!} e^{k} =e^{e}-1