$\begin{align}  & {{2}^{n}}>n \\ & \text{When }n=1 \\ & {{2}^{1}}>1 \\ & \text{when }n=2 \\ & {{2}^{2}}>2 \\ & \text{The formula is valid for }n \\ & \text{Let }n=k \\ & {{2}^{k}}>k \\ & \text{For }k+1 \\ & {{2}^{k+1}}={{2}^{k}}\cdot 2>2k\text{   (by assumption} \\ & {{2}^{k+1}}>2k \\ & Since\text{ }2k=k+k>k+1\text{ for all }k>1,\text{ it follows that} \\ & {{2}^{k+1}}>2k>k+1 \\ & \text{Hence, }{{2}^{n}}>n\text{ for all integers }n\ge 1 \\\end{align}$