How do you get $\displaystyle \frac{1}{n!}\leq\frac{1}{2^{n-1}}$ for $\displaystyle n>1$?
$\displaystyle \displaystyle n \cdot (n-1) \cdot (n-2) \cdot \dots \cdot 3 \cdot 2 > 2 \cdot 2 \cdot 2 \cdot \dots \cdot 2 \cdot 2$ (with $\displaystyle \displaystyle n-1$ terms on each side of the inequality)
$\displaystyle \displaystyle n! > 2^{n-1}$
$\displaystyle \displaystyle \frac{1}{n!} < \frac{1}{2^{n-1}}$.
Fpr those who didn't "get it", both n! and $\displaystyle 2^{n-1}$ have n- 1 factors (not counting the "1" in n!). Just compare the size of the factors. n! is the product of all numbers from n down to 2 while $\displaystyle 2^{n-1}$ has all "2"s. 2= 2 while 3> 2 so 3(2)> 2(2), 4> 2 so 4(3)(2)> 2(2)(2), 5> 2 so 5(4)(3)(2)> 2(2)(2)(2), etc.