\begin{equation} \left(1+\frac{1}{n}\right)^n = 1 + \sum_{k=1}^n\left\{\frac{1}{k!}\prod_{r=0}^{k-1}\left(1-\frac{r}{n}\right)\right\}\,.\tag{1} \end{equation}
Proof.
By the binomial theorem
$$\left(1+\frac{1}{n}\right)^n = 1+ \sum_{k=1}^n {{n}\choose{k}}\left(\frac{1}{n}\right)^{k}.$$ Hence, to prove $(1)$, we must prove that
$$ \sum_{k=1}^n {{n}\choose{k}}\left(\frac{1}{n}\right)^{k} = \sum_{k=1}^n\left\{\frac{1}{k!}\prod_{r=0}^{k-1}\left(1-\frac{r}{n}\right)\right\}\,,\tag{2} $$
for all positive integers $n$.
At present I can't find any reasonable quantity that if added on both sides leads to equality for $n+1$, i.e. the quantity that completes the proof. Any hint is appreciated.
EDIT: A similar question has been asked in this and this posts. What I'd like to do is to prove that (1) (or (2)) holds for any integer $n>0$. @Netchaiev's answer and those provided in the aforementioned posts are correct for a specific $n$, but what about for an arbitrary integer $n>0$ ?
I guess that the proof-by-induction should do the job but I don't see how.
You have $$ \begin{align*} \frac{1}{k!}\prod_{r=0}^{k-1}\left(1-\frac{r}{n}\right)& = \frac{1}{k!}\prod_{r=0}^{k-1}\left(\frac{n-r}{n}\right) \\ & = \frac{1}{k!}\frac{n\times \cdots \times (n-k+1)}{n^k} \\ & = \frac{n!}{k!(n-k)!}\frac{1}{n^k} \\ & = \binom{n}{k}\left(\frac{1}{n}\right)^k \end{align*} $$ and the global equality follows.