Question

A) Basis Step
b) Inductive Step

Answer 1

Basis step: Let
`n=2`. Then

`2! = 2\cdot1 = 2`

and

`2^2 = 4`

so
`2! < 2^2`.

Inductive step: Assume
`k! < k^k`. We want to show that
`(k+1)! < (k+1)^(k+1)`.

Now

`(k+1)! = (k+1) \cdot k! \\\\ ~~~~~~~~~~~ < (k+1) \cdot k^k \\\\ ~~~~~~~~~~~ < (k+1) \cdot k^(k+1) \\\\ ~~~~~~~~~~~ < (k+1) \cdot (k+1)^(k+1)`

where the first inequality follows from the induction hypothesis; the second follows from multiplying the right side by some
`k>2`; and the third from adding the remaining terms to complete the binomial expansion
`(k+1)^(k+1)`, all of which are positive since they are some power of
`k>2`.

QED