Can anyone help me confirm if I've got this one correct? Particularly the part at

**A**, where I've defined (k + 2)! = (k + 1)!(k + 2).

Many thanks.

**Q.** (n + 1)!

__>__ 2

^{n}, for n

**Attempt:** __Step 1:__ For n = 1...

(1 + 1)! = 2! = 2 & 2

^{1} = 2

Since 2

__>__ 2, the statement is true for n = 1.

__Step2:__ Assume the statement is true for n = k, i.e. assume (k + 1)!

__>__ 2

^{k}.

We must now show that the statement is true for n = k + 1,

i.e. ((k + 1) + 1)!

__>__ 2

^{k+1} if (k + 2)!

__>__ 2

^{k+1} **A:** (k + 2)! = (k + 1)!(k + 2)

__>__ (k + 1)2

^{k}...((k + 1)!

__>__ 2

^{k}...assumed)

If (k + 1)2

^{k} __>__ 2

^{k+1}, then (k + 2)!

__>__ 2

^{k+1}
if (k + 1)2

^{k} __>__ 2

^{k+1}
if (k + 1)2

^{k} - 2

^{k+1} __>__ 0

if 2

^{k}[(k + 1) - 2]

__>__ 0...which is true for k

__>__ 1

=> (k + 2)!

__>__ 2

^{k+1}
Therefore, the statement is true for n = k + 1, if true for n = k. Thus, the statement is true for all n

__>__ 2, n