I have a hard time to understand a certain part of the proof below
Let $e_n:=(1+\frac{1}{n})^n$ and $s_n:=\sum_{v=0}^{n}\frac{x^v}{v!}$
According to the binomial Theorem we have
$$e_n=1+n\cdot\frac{1}{n}+\frac{n(n-1)}{2!}\cdot\frac{1}{n^2}+...+\frac{n(n-1)(n-2)...1}{n^n}$$
$$=1+1+\frac{1}{2!}(1-\frac{1}{n})+...+\frac{1}{n!}(1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{n-1}{n})$$
$$<s_n(1)$$
this gives us
$$e=\lim_{n\rightarrow\infty}e_n\leq\lim_{n\rightarrow\infty}s_n(1)=\exp(1)$$
What is the justification for the $\leq$ symbol?
I know the limits exists, the left Limit was proved by constructing nested intervals $I_n=[(1+\frac{1}{n})^n,(1+\frac{1}{n})^{n+1}]$, the right Limit exists because of the quotient criterion.
The proof continues with:
Let $k>n$
Then
$$e_k>1+1+\frac{1}{2!}(1-\frac{1}{k})+...+\frac{1}{n!}(1-\frac{1}{k})(1-\frac{2}{k})...(1-\frac{n-1}{k})(*)$$
If we fixate a $n$ and let $k$ approach infinity we get $e\geq s_n(1)$ (Why?) which gives us
$$e\geq \lim_{n\rightarrow\infty}s_n(1)$$
What was the logic of fixating a $k$ and why is it still the same Limit after the Fixation and why are we allowed to do that ?
Please shed some light on those equations.