#### Answer

See the full explanation below.

#### Work Step by Step

Let ${{S}_{n}}$ ${{2}^{n}}>{{n}^{2}}$ for $ n\ge 5$
Let ${{S}_{n}}$ be a statement involving the positive integer n. If
1. ${{S}_{5}}$ is true, and
2. Let us assume that the ${{S}_{k}}$ be true for some integer $ k $.
Now, one has to use ${{S}_{k}}$ to prove ${{S}_{k+1}}$ is also true. If one is able to prove ${{S}_{k+1}}$ then by the principle of mathematical induction the statement ${{S}_{n}}$ is true for all positive integers $ n\ge 5$.
Let ${{S}_{n}}:{{2}^{n}}>{{n}^{2}}$
For, ${{S}_{5}}$ put $ n=5$ in the equation;
$\left( {{2}^{5}} \right)>\left( {{5}^{2}} \right)$
By simplifying, one gets
$32>25$
Now, one has satisfied the first equation.
Going to the second condition, assume that $ S\left( n \right)$ is true for $ n=k $ and prove that the statement is true for $ n=k+1$.
${{S}_{k}}:{{2}^{k}}>{{k}^{2}}$ (I)
And
${{S}_{k+1}}:{{2}^{k+1}}>{{\left( k+1 \right)}^{2}}$
${{S}_{k+1}}:{{2}^{k+1}}>{{\left( k+1 \right)}^{2}}$
Simplify the terms
$\begin{align}
& {{2}^{k+1}}=2\left( {{2}^{k}} \right) \\
& >2\left( {{k}^{2}} \right) \\
& ={{k}^{2}}+{{k}^{2}}
\end{align}$
Now, ${{k}^{2}}>2k+1$ for all the values of $ k>5$
So, the above equation can be written as
$\begin{align}
& {{2}^{k+1}}=2({{2}^{k}}) \\
& >2\left( {{k}^{2}} \right) \\
& ={{k}^{2}}+{{k}^{2}}
\end{align}$
$\begin{align}
& >{{k}^{2}}+2k+1 \\
& >{{\left( k+1 \right)}^{2}}
\end{align}$
This inequality for $ k+1$ is the same as the previous one which means it has satisfied the second condition of mathematical induction.
Therefore, by the method of extended mathematical induction, it is proved that ${{2}^{n}}>{{n}^{2}}$ for $ n\ge 5$.