Mathematical induction is the way to go, but first you want to have a "target." I'm not sure what you did, but I think you confused $c$ and $d$ somewhere along the line in your calculations. Let's see a first few values:
$\begin{align*} A&= \left(\begin{array}{cc}1&c\\0&d\end{array}\right)\\ A^2 &= \left(\begin{array}{cc}1&c\\0&d\end{array}\right)\left(\begin{array}{cc}1&c\\0&d\end{array}\right) = \left(\begin{array}{cc}1 & c+cd\\0&d^2\end{array}\right).\\ A^3 &= \left(\begin{array}{cc} 1&c+cd\\ 0 &d^2\end{array}\right) \left(\begin{array}{cc} 1 & c\\ 0 & d\end{array}\right) = \left(\begin{array}{cc} 1 & c+cd+cd^2\\ 0 & d^3 \end{array}\right).\\ A^4 &= \left(\begin{array}{cc} 1&c+cd+cd^2\\ 0 & d^3 \end{array}\right)\left(\begin{array}{cc} 1 & c\\0 & d\end{array}\right) = \left(\begin{array}{cc} 1 & c + cd + cd^2 + cd^3\\ 0 & d^4 \end{array}\right). \end{align*}$ Okay, that suggests the pattern:
Conjecture. For every positive integer $n$, $A^n = \left(\begin{array}{cc} 1 & c(1+d+d^2+\cdots + d^{n-1})\\ 0 & d^n \end{array}\right).$
To prove the conjecture, we use mathematical induction. Prove the formula is true for $n=1$, and then, assuming the formula holds for $k$, prove it holds for $k+1$. We have:
Basis. For $n=1$, we have $A = \left(\begin{array}{cc}1 & c\\0 & d\end{array}\right),$ so the formula holds.
Inductive step. Show that if the formula holds for $k$, then it also holds for $k+1$.
Induction hypothesis. Assume the result holds for $k$; that is, that $A^k = \left(\begin{array}{cc} 1 & c(1+d+\cdots+d^{k-1})\\0 & d^k \end{array}\right).$
Now we have: $\begin{align*} A^{k+1} & = A^kA\\ &= \left(\begin{array}{cc} 1 & c(1+d+\cdots + d^{k-1})\\0 & d^k\end{array}\right) \left(\begin{array}{cc} 1 & c\\0 & d\end{array}\right)\\ &= \left(\begin{array}{cc} 1 & c + cd(1+d+\cdots + d^{k-1})\\ 0 & d^{k+1}\end{array}\right)\\ &= \left(\begin{array}{cc} 1 & c(1 + d(1+d+\cdots + d^{k-1}))\\ 0 & d^{k+1}\end{array}\right)\\ &= \left(\begin{array}{cc} 1 & c(1+d+d^2+\cdots + d^k)\\ 0 & d^{k+1}\end{array}\right). \end{align*}$ That is exactly the formula we have evaluated at $k+1$. Therefore, if the formula holds for $k$, then it holds for $k+1$ as well.
So: the formula holds for $n=1$; and if it holds for $n=k$, then it also holds for $n=k+1$. By Mathematical Induction, we conclude that the formula holds for all $n$.