Comparison test to prove $\frac{1}{p!}<\frac{1}{2^{p-1}}$ I'm trying to show $\frac{1}{p!}<\frac{1}{2^{p-1}}$ for $p>2$
I notice that the right hand side of the inequality represents a geometric sequence. I can compute its infinite sum but how do I show the left hand side is smaller?
 A: Computing the geometric sum $\sum_{p=1}^\infty \frac{1}{2^{p-1}}$ is irrelevant to proving this inequality. Note that it is equivalent to proving $$2^{p-1}<p!$$ for $p>2$. You should prove this inequality with induction. The base case of $p=3$ should be straightforward. For the induction step, you can multiply the inequality  $2^{k-1}<k!$ by $2$, and then use the fact that $2<k+1$ to quickly see that $2^k< (k+1)!$. Once you've done that, you know the inequality will hold for all $k \in \Bbb{N}$ where $k>3$, and you can flip the inequality back to get the result you are after.
A: Prove by induction the equivalent statement $n!>2^{n-1}$ for $n>2$.

First, show that this is true for $n=3$:
$3!>2^{3-1}$
Second, assume that this is true for $n$:
$n!>2^{n-1}$
Third, prove that this is true for $n+1$:
$(n+1)!=n!\cdot(n+1)>\color{red}{n!}\cdot2>\color{red}{2^{n-1}}\cdot2=2^n$

Please note that the assumption is used only in the part marked red.
A: hint: $2^{p-1} = 2\cdot 2\cdots2 < 2\cdot 3\cdot (2\cdot 2) \cdot 5\cdot 6\cdots p$
