Simplifying sum of falling factorials
You gave the closed form for the number of operations:
$$
\sum_{m=1}^n n(n-1)\cdots(n-m+1) = \sum_{m=1}^n \frac{n!}{(n-m)!} = n! \sum_{m=0}^{n-1} \frac{1}{m!}.
$$
We can estimate this sum as follows:
$$
n! \sum_{m=0}^{n-1} \frac{1}{m!} =
n! \left(e - \sum_{m=n}^\infty \frac{1}{m!}\right) = 
en! - 1 - \frac{1}{n+1} - \frac{1}{(n+1)(n+2)} - \cdots = \\
en! - 1 - O\left(\frac{1}{n}\right).
$$