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**GrigOrig99** 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 $\displaystyle \in \mathbb{N}$

**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 $\displaystyle \in \mathbb{N}$