I have this messy proof of $\lim_{n \to \infty}(1+\frac{1}{n})^n=e$ in my notebook. I can't find it anywhere else, but I need it since the professor accepts only this version at the exam. At the time I am only stuck with the first part, so I will write the proof only to that point. Also, there is a few steps with missing reasoning between them.
Proof.
We are given two sequences:
$$x_n=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\dots+\frac{1}{n!}$$ $$y_n=\left(1+\frac{1}{n}\right)^n$$
If $x_n$ converges, then $\lim_{n \to \infty}x_n=e$. (This is given.)
Let's prove that $x_n$ converges.
To prove that, the Monotone Convergence Theorem will be used.
Since $x_{n+1}>x_{n}$ the sequence is increasing. (I skipped the unnecessary reasoning for this.)
Now let's prove that it is bounded from above.
$$n!=1\cdot 2\cdot 3\cdots n>2^{n-1}\tag{1.1}$$ $$\frac{1}{n!}\leq \frac{1}{2^{n-1}}\tag{1.2}$$
$$\underbrace{2}_{\text{?}}\leq x_n\leq 1+1+\frac{1}{2}+\frac{1}{2^2}+\dots+\frac{1}{2^{n-1}}\tag{1.3}$$
$$1+q+q^2+\dots+q^{n-1}=\underbrace{\overbrace{\frac{1-q^n}{1-q}}^{?}=\frac{q^n-1}{q-1}}_{\text{?}}\tag{1.4}$$
$$q=\frac{1}{2}$$
$$\underbrace{1}_{\text{?}}+\frac{1-\frac{1}{q^n}}{1-\frac{1}{q}}=1+2\left(1-\frac{1}{2^n}\right)<1+2=3\tag{1.5}$$
And so $2\leq x_n\leq3$, $x_n$ is bounded and increasing, hence it converges. Therefore $\lim_{n \to \infty}x_n=e$.
The proof proceeds with proving that $\lim_{n\to \infty}y_n=Y$, and then showing $y\geq e$ and $y \leq e$, and thereby $y=e$. This part is very long and already looking messy in the notebook. Because of that I will skip it, since it isn't important for the question.
My questions:
Is there any obvious reason why $2^{n-1}$ is a good (necessary?) choice for the inequality (1.1)?
How does (1.2) follow from (1.1)? Why is it $\leq$ and not $ < $ ? I see that $\leq$ is the right choice since for $n=1$, we have $\frac{1}{1!}=\frac{1}{2^{1-1}}$, but is there any way to know this without evaluating for choices of $n$?
Why is that $2$ there in inequality (1.3)? Where did it come from ?
Considering inequality (1.4) there is few things;
a) How was the overbraced fraction derived from the geometric series?
b) The underbraced equality at first didn't make sense to me, but when I evaluated for few $n$s, I realized it actually holds. Is this a known thing, that the order of subracting isn't important for two fractions to be equal, as long as both the numerator and the denominator of the fraction are both $<0$ or $>0$? Also what was the role of the equality in the proof, or was it just a remark ?
EDIT: I just realized that we can do that since the fraction is always positive. But it still doesn't seem so obvious to notice. Should it be obvious?
Why is that $1$ there in the inequality (1.5) ? Where did it come from ?
If you have a source for this proof please send me a link or tell me where I can find it.
Thanks