My friend needs help and requires this work to be done by today. I hope it won't trouble anyone to elaborate how to do so.
EDIT: Where it says $\displaystyle 2^{2}$, it's supposed to be $\displaystyle 2^{N}$ I apologies.
$\displaystyle \displaystyle
k<2^k \ \mbox{multiple by 2}\Rightarrow \ 2k<2*2^k\Rightarrow k+k<2^{k+1}$
Since $\displaystyle \displaystyle k\geq n\geq 1$, we can take the 2nd k and set it equal to 1 without loss of generality.
$\displaystyle \displaystyle k+1<2^{k+1}$
P(k)
$\displaystyle k<2^k$
P(k+1)
Replace k with k+1
$\displaystyle k+1<2^{k+1}\;\;\;?$
Proof
$\displaystyle k<2^k\Rightarrow\ 2(k)<2\left(2^k\right)$
$\displaystyle 2k<2^{k+1}$
$\displaystyle 2k=k+k\Rightarrow\ (k+k)<2^{k+1}$
Since it is proven for the base case k=0 or k=1,
then $\displaystyle (k+1)<(k+k)$ for $\displaystyle k>1$
Therefore, if $\displaystyle k<2^k$
then $\displaystyle (k+1)<2^{k+1}$ since $\displaystyle (k+1)<(k+k)$ for $\displaystyle k>1$