Given, If $A=\left[\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right]$, then $A^n=\left[\begin{array}{ll}1 & 0 \\ n & 1\end{array}\right], \forall n \in N$
We will generalise the exponent of $A$.
Let $n=2$
$$
\begin{aligned}
\therefore A^2 & =\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right] \cdot\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right] \\
& =\left[\begin{array}{cc}
1 \cdot 1+0 \cdot 1 & 1 \cdot 0+0 \cdot 1 \\
1 \cdot 1+1 \cdot 1 & 1 \cdot 0+1 \cdot 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right]
\end{aligned}
$$

Again, let $n=3$
$$
\therefore A^3=A^2 \cdot A=\left[\begin{array}{ll}
1 & 0 \\
2 & 1
\end{array}\right] \cdot\left[\begin{array}{ll}
1 & 0 \\
1 & 1
\end{array}\right]=\left[\begin{array}{cc}
1 & 0 \\
2+1 & 1
\end{array}\right]=\left[\begin{array}{ll}
1 & 0 \\
3 & 1
\end{array}\right]
$$

Similarly, $A^4=\left[\begin{array}{ll}1 & 0 \\ 4 & 1\end{array}\right]$
Hence, $A^n=\left[\begin{array}{ll}1 & 0 \\ n & 1\end{array}\right]$ is true for $n=3$