Can anyone help confirm if I have solved this correctly. Specifically the part preceded byA, where I have defined the coefficient as 2^{k}. I had originally thought that it might to be 2^{k-1}, but subbing 3 into that version yielded negative answers.

Many thanks.

n! > 2

Q.^{n-1}for n>3, n $\displaystyle \in \mathbb{N}$

Attempt:Step 1:For n = 3...

3! = 1 x 2 x 3 = 6 & 2^{3-1 }= 4

Since 6 > 4, the statement is true for n = 3.

Step 2:Assume the statement is true for n = k, i.e. assume k! > 2^{k-1}

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

i.e. (k + 1)! > 2^{(k+1)}^{-1}= (k + 1)! > 2^{k}

(k + 1)! = (k + 1)k! > (k + 1)2^{k-1}...(k! > 2^{k-1}...assumed)

If it can be shown that (k + 1)2^{k-1}> 2^{(k+1)-1}, then (k + 1)! > 2^{k}

If (k + 1)2^{k-1}> 2^{k}

If (k + 1)2^{k-1}- 2^{k}> 0

A:If 2^{k}[(k + 1)2^{k-2}- 1] > 0...true when>3

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

Therefore, the statement is true for n = k + 1, if true for n = k. Thus the statement is true for all n>3, n $\displaystyle \in \mathbb{N}$