<s>[INST] Proving sequence equality using the binomial theorem The problem: Prove that for $n \in \mathbb N$:
$$ \left(1 + \frac{1}{n} \right)^n = 1 + \sum_{m=1}^{n} \frac{1}{m!} \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right). $$
The hint is to use the binomial theorem. So the left side can become:
$$ \sum_{m=0}^{n} \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m $$
I don't really know where to go from here, I've tried manipulating the expressions to make them look similar but I'm not really getting anywhere.
 [/INST] Take your second sum
$$
\sum_{m=0}^{n} \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m
$$
and write it as 
$$
1+\sum_{m=1}^n  \frac{n!}{m!(n - m)!} \left(\frac{1}{n} \right)^m
$$
to get the indices to match.
In your first sum
$$
\sum_{m=1}^{n} \frac{1}{m!} \left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right)
$$
ignoring the $\frac{1}{m!}$ for now, notice
$$
\left(1 - \frac{1}{n} \right) \left(1 - \frac{2}{n} \right) \cdots \left(1 - \frac{m-1}{n} \right)=\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\cdots\left(\frac{n-m+1}{n}\right).
$$
Multiplying by $1=\frac{n}{n}$ gives
$$
\frac{n}{n}\left(\frac{n-1}{n}\right)\left(\frac{n-2}{n}\right)\cdots\left(\frac{n-m+1}{n}\right)=\dots
$$
Can you take it from there?
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