Answer
See the explanation below.
Work Step by Step
We use mathematical induction as follows:
Statement ${{S}_{1}}$ is:
${{2}^{n-1}}=1$
And simplifying on the right, we obtain ${{2}^{n}}-1=1$.
So, this true statement shows that ${{S}_{1}}$ is true.
And assume ${{S}_{k}}$ is true. Using ${{S}_{k}},{{S}_{k+1}}$ from the expression,
$\begin{align}
& {{S}_{k}}=1+2+{{2}^{2}}+\ldots +{{2}^{k-1}} \\
& ={{2}^{k}}-1
\end{align}$
Adding ${{2}^{\left( k+1 \right)-1}}$ on both sides as:
$\begin{align}
& 1+2+{{2}^{2}}+\ldots +{{2}^{\left( k+1 \right)-1}}={{2}^{k}}-1+{{2}^{\left( k+1 \right)-1}} \\
& 1+2+{{2}^{2}}+\ldots +{{2}^{\left( k+1 \right)-1}}={{2}^{k}}-1+{{2}^{\left( k \right)}} \\
& 1+2+{{2}^{2}}+\ldots +{{2}^{\left( k+1 \right)-1}}=2\times {{2}^{k}}-1 \\
& 1+2+{{2}^{2}}+\ldots +{{2}^{\left( k+1 \right)-1}}={{2}^{k+1}}-1
\end{align}$
Thus, ${{S}_{k+1}}$ is true. The result ${{S}_{n}}=1+2+{{2}^{2}}+\ldots +{{2}^{n-1}}={{2}^{n}}-1$ holds true.