Prove that $(1+\frac{1}{n}{)}^n\le e<3$.

Using Newton's formula (also called the Binomial Theorem)

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{{(1+\frac{1}{n})}^n}& =& 1+n\left(\frac{1}{n}\right)+\frac{n(n-1)}{2!}\left(\frac{1}{n^2}\right)+\frac{n(n-1)(n-2)}{3!}\left(\frac{1}{n^3}\right)+\dots \\ \multicolumn{1}{c}{}& \le & 1+1+\frac{1}{2!}+\frac{1}{3!}+\dots \\ \multicolumn{1}{c}{}& \le & e<3.\end{array}}$

(a) Which is larger: $1.{000001}^{1000000}$ or 2?

It's possible to prove that every number in the form $(1+\frac{1}{a}{)}^a$ is greater than 2.

${\displaystyle {(1+\frac{1}{a})}^a=1+a·\frac{1}{a}+\text{other positive terms}>2}$

for every integer a> 2. For a=1000000, the problem is solved.

(b) Which is larger: ${100}^{300}$ or or 300! (i.e.factorial 300)?

This is a bit more complex. I'll use the formula $(1+\frac{1}{n}{)}^n=(\frac{n+1}{n}{)}^n<3\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}}(1)$

I will now use the last inequality to prove by induction that $n!>(\frac{n}{3}{)}^n.$ Clearly this is true for $n=1$ and 2 and so using the induction hypothesis that it is true for $n$:

${\displaystyle \begin{array}{ccc}\multicolumn{1}{c}{(n+1)!=(n+1)n!}& \le & (n+1){\left(\frac{n}{3}\right)}^n\\ \multicolumn{1}{c}{}& =& 3{\left(\frac{n+1}{3}\right)}^{n+1}{\left(\frac{n}{n+1}\right)}^n.\\ \multicolumn{1}{c}{}\end{array}}$

So using (1) gives:

${\displaystyle (n+1)!\le {\left(\frac{n+1}{3}\right)}^{n+1}.}$

The demonstration by induction is complete. In particular, for n=300, the formula solves the given problem.