#### Answer

See the full explanation below.

#### Work Step by Step

Let ${{S}_{n}}$ be a statement involving the positive integer n. If
1. ${{S}_{1}}$ is true, and
2. Let us assume that the ${{S}_{k}}$ be true for some integer $ k $.
Then, 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 $.
Firstly, show that ${{S}_{1}}$ is true.
For ${{S}_{1}}$ one has
And write ${{S}_{1}}$ by taking the its first term on left and substituting n with $1$ on the right.
$\begin{align}
& {{S}_{1}}: \\
& {{\left( \frac{a}{b} \right)}^{1}}=\frac{{{a}^{1}}}{{{b}^{1}}} \\
& =\frac{a}{b}
\end{align}$
Therefore, the statement ${{S}_{1}}$ is true.
Then, let us assume that ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ should also be true by the method of mathematical induction.
${{S}_{k}}:{{\left( \frac{a}{b} \right)}^{k}}=\frac{{{a}^{k}}}{{{b}^{k}}}$
Now, one has to prove ${{S}_{k+1}}$ is true
${{S}_{k+1}}:{{\left( \frac{a}{b} \right)}^{k+1}}=\frac{{{a}^{k+1}}}{{{b}^{k+1}}}$
Since, one assume that ${{S}_{k}}$ is true,
${{\left( \frac{a}{b} \right)}^{k}}=\frac{{{a}^{k}}}{{{b}^{k}}}$
And multiply $\left( \frac{a}{b} \right)$ on the both sides, $\begin{align}
& {{\left( \frac{a}{b} \right)}^{k}}\left( \frac{a}{b} \right)=\frac{{{a}^{k}}}{{{b}^{k}}}\left( \frac{a}{b} \right) \\
& {{\left( \frac{a}{b} \right)}^{k+1}}=\frac{{{a}^{k+1}}}{{{b}^{k+1}}} \\
\end{align}$
So, the final statement is ${{S}_{k+1}}$.
Thus, by the method of principal of mathematical induction, the statement ${{\left( \frac{a}{b} \right)}^{n}}=\frac{{{a}^{n}}}{{{b}^{n}}}$ is true for every positive integer $ n $.