Calculating ...5(5+4(4+3(3+2(2+1(1))))) I've been puzzling over this question for a while now, and I've finally decided to turn to the StackExchange community in order to get an answer.  
How would one determine the value of the expression ...5(5+4(4+3(3+2(2+1(1)))))) to n, assuming that in this case n = 5?  
Cheers!

This can be written as:
$$5\cdot 5 + 5\cdot 4\cdot 4 + 5\cdot 4\cdot 3\cdot3 + 5\cdot 4\cdot 3\cdot 2\cdot 2+5!\cdot 1\\
=\frac{5!}{4!}\cdot 5 + \frac{5!}{3!}\cdot 4 + \frac{5!}{2!}\cdot 3 + \frac{5!}{1!}\cdot 2+\frac{5!}{0!}\cdot 1$$
Then the general formula is:
$$\sum_{k=0}^{n-1}\frac{n!}{k!}(k+1) = n!\sum_{k=0}^{n-1}\frac{k+1}{k!}$$
This likely does not have a closed form, but we can say that:
$$\sum_{k=0}^{n-1}\frac{k+1}{k!} = \frac{1}{(n-1)!}+2\sum_{k=0}^{n-2}\frac{1}{k!} $$
So:
$$\lim_{n\to\infty}\sum_{k=1}^{n-1}\frac{k+1}{k!}=2e$$
