$$\left(1+\frac 1n\right)^{n} = 1 + \sum\limits_{k=1}^n \left\{\frac 1{k!}\prod_{r=0}^{k-1}\left(1-\frac rn\right)\right\}$$
this exercise is taken from Apostol's Calculus I (page 45) and it's supposed to be proved by using the binomial theorem. I can only carry out the expression on the left this way:
$$\left(1+\frac 1n\right)^{n} = \sum_{k=0}^n{n \choose k}1^k\left(\frac 1n\right)^{n-k}=1 + \sum_{k=0}^{n-1}{n \choose k}\frac 1{n^{n-k}}$$
I have no clue on how to proceed.. any help is very appreciated, thanks in advance!
I think you can continue from there: $$\begin{align}{n \choose k}\frac 1{n^{n-k}}&=\frac{n(n-1)(n-2)\cdots (n-k+1)}{k!n^{n-k}}\\ &=\frac{1}{k!n^{n-k}}\prod_{r=0}^{k-1}(n-r)\\ &=\frac{1}{k!}\prod_{r=0}^{k-1}\frac{n-r}{n}\end{align}$$
and you can make the final step!. But for the indexes for the sum in binomial theorem you might write: $$(1+\frac 1n)^{n} = \sum_{k=0}^n{n \choose k}1^k(\frac 1n)^{n-k}=1 + \sum_{k=1}^{n}{n \choose k}\frac 1{n^{n-k}}$$ (because as you know ${n \choose 0}=1$ and what I wrote in the product above is not be true id $k=0$ because in this case we will have $\prod_{r=1}^{0-1}$ which does not have a sense)