Use proof by induction to prove $ \frac{1}{n!}<\frac{1}{2^n-1} $ for all $n\geq 4$ Use proof by induction to prove that  that  $ \frac{1}{n!}<\frac{1}{2^n-1} $ for all $n\geq 4$, .\Base case: $$\frac{1}{4}=\frac{1}{24}\leq \frac{1}{2^4-1}$$
Inductive hypothesis: Assume there exists $k\in \mathbb{N}$  s.t.
$$ \frac{1}{k!}\leq\frac{1}{2^k-1} $$
Inductive step: Show that:$$ \frac{1}{(k+1)!}\leq\frac{1}{2^{k+1}-1} $$
Now, $$\frac{1}{(k+1)!}=\frac{1}{k!}\cdot\frac{1}{k+1}$$
Using the hypothesis $$\frac{1}{(k+1)!}\leq\frac{1}{2^k-1}\cdot \frac{1}{k+1}$$
Because $n\geq4, \frac{1}{k+1}<\frac{1}{2}$
$$\frac{1}{(k+1)!}\leq\frac{1}{2^k-1}\cdot \frac{1}{2}=\frac{1}{2^{k+1}-2}\leq\frac{1}{2^{k+1}-1}$$
Hence by mathematical induction we have proved that $ \frac{1}{n!}<\frac{1}{2^n-1} $ for all $n\geq 4$
Firstly I need to know if the proof is correct, secondly it has to be as concise as possible hence I would like to know if there are any lines I can change/delete
And lastly can anyone explain to me why every sentence starts with "\" It looks perfectly fine in www.sharelatex.com ;(

Rewrite as$$n!\ge2^n.$$
Then
$$4!\ge2^4$$
and
$$n!\ge2^n\land n+1\ge2\implies(n+1)!\ge2^{n+1}.$$
