Bounding Euler's e by showing $2 \leq \left (1+ \frac{1}{n}\right)^n < 3$ 
(a)For $m,n \in \mathbb{N}$, $m<n$ and $k=0, ..., m$:
$$\frac{1}{m^k} \binom{m}{k} \leq \frac{1}{n^k} \binom{n}{k}$$
(b)For $n\in \mathbb{N}$ and $k=1, ..., n:$
$$\frac{1}{n^k}  \binom{n}{k} \leq \frac{1}{k!} \leq \frac{1}{2^{k-1}}$$
(c)Show that for $n \in \mathbb{N}:$
$$2 \leq (1+ \frac{1}{n})^n < 3$$
Use (a) and the binomial theorem for (b). Use the following for (c): $$\sum_{j=0}^{n}q^j=\frac{1-q^{n+1}}{1-q}$$

I've already proven (a) thanks to @trancelocation. Now I got stuck on (b) and (c).
(b) I don't know how to use (a) and the binomial theorem here.
$\frac{1}{n^k}  \binom{n}{k} \leq \frac{1}{k!} \leq \frac{1}{2^{k-1}} \Leftrightarrow \frac{n!}{n^k*(n-k)!* k!}\leq \frac{1}{k!} \leq \frac{1}{2^{k-1}} \Leftrightarrow \frac{n!}{n^k*(n-k)!}\leq 1 \leq \frac{k!}{2^{k-1}}$
Now I have to show: $\frac{n!}{n^k*(n-k)!}\leq 1$ and $1 \leq \frac{k!}{2^{k-1}}$
$\frac{n!}{n^k*(n-k)!} = \frac{n*(n-1)*...*(n-k+1)*[(n-k)*...*1]}{n^k*[(n-k)*...*1]}=\frac{n*...*(n-k+1)}{n^k}=\frac{n}{n}*\frac{n-1}{n}*...*\frac{n-k+1}{n} \leq 1$.
The first factor is $1$. Starting from the second factor their value gets smaller, so the entire term gets smaller then $1$.
I tried to do it like this:
for $i=0, ..., n$:  $\frac{n-i}{n} \leq 1 \Leftrightarrow n-i \leq n \Leftrightarrow 0 \leq i$
For the second part:
$\frac{k!}{2^{k-1}} \geq 1 \Leftrightarrow \frac{k!}{2^k*2} \geq 1 \Leftrightarrow \frac{k!}{2^k} \geq 2\Leftrightarrow \frac{k}{2}*\frac{k-1}{2}*\frac{k-2}{2}*...*\frac{1}{2} \geq 2$
So for $\frac{k-j}{2} \geq 2$, with $j=0, ..., k-1$
$\frac{k-j}{2} \geq 2 \Leftrightarrow k-j \geq 4$
This doesn't seem to help at all.
(c)$2 \leq (1+ \frac{1}{n})^n < 3$
I didn't know how to start at c) at all. The provided formula doesn't help me.
 A: For the part (b) you had the right idea but you made a little mistake. Here is the correct one:
$\frac{k!}{2^{k-1}} \geq 1 \Leftrightarrow \frac{2 \cdot k!}{2^k} \geq 1 \Leftrightarrow \frac{k!}{2^k} \geq \frac12\Leftrightarrow \frac{k}{2}*\frac{k-1}{2}*\frac{k-2}{2}*...*\frac{1}{2} \geq \frac12$
or
$\frac{k}{2}*\frac{k-1}{2}*\frac{k-2}{2}*...*\frac{3}{2}*\frac{2}{2} 
\geq 1$
and since all factors on the LHS are $≥1$ this is true. 
For part (c), we need to solve two inequalities. First, let us rewrite
the geometric series $\sum_{j=0}^{n}q^j=\frac{1-q^{n+1}}{1-q}$ as $q^n = \frac1q (1 - (1-q)\sum_{j=0}^{n}q^j)$. Using $q = 1 + \frac1n$ and $(1 + \frac1n)^j \ge 1$ gives 
$$
(1 + \frac1n)^n = \frac{1}{1 + \frac1n} (1 +\frac1n\sum_{j=0}^{n}(1 + \frac1n)^j) \\
= \frac{1}{1 + \frac1n} (1 +\frac1n+\frac1n\sum_{j=1}^{n}(1 + \frac1n)^j)
= \frac{1 +\frac1n}{1 + \frac1n}  +\frac1n\sum_{j=1}^{n}\frac{(1 + \frac1n)^j}{1 + \frac1n}\\
= 1 + \frac1n\sum_{j=0}^{n-1}(1 + \frac1n)^j \ge 1 + \frac{n}{n} =2
$$
which establishes the left inequality.  For the right inequality, use the binomial theorem, then (b), then the given geometric series, to write
$$ 
(1 + \frac1n)^n = \sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k} = 1  + \sum_{k=1}^{n} \binom{n}{k} \frac{1}{n^k}\\
\le 1  + \sum_{k=1}^{n} \frac{1}{2^{k-1}} =  1  + \sum_{k=0}^{n-1} \frac{1}{2^{k}}  = 1 + \frac{1 - (\frac12)^n}{1 - \frac12} < 1 + \frac{1 }{1 - \frac12} = 3
$$
