Question

Show that 2^(n-1) <= n! for all n in the set of natural numbers.

Initial step: The statement is true for n = 1. 2^(1-1) = 2^0 = 1, and 1! = 1. Hence, 2^(n-1) <= n! holds for n = 1.

Inductive step: Assume the statement is true for some k in the set of natural numbers.

This means that 2^(k-1) <= k! for some k in the set of natural numbers.

Then, for n = k + 1:

2^((k + 1) - 1) = 2^k
= 2(2^(k-1))
<= 2(k!), by the inductive hypothesis
<= (k+1)k!, as 2 <= k + 1 (where does the 2 <= k + 1 come from?)
= (k+1)!

Answer 1

k starts at 1

Answer 2

your initial case was n=1 ( ie k=1)

Answer 3

Makes sense, thanks.