Finding $n$th power of a $3\times 3$ matrix Find the $A^n$ if $$A=\begin{bmatrix}1 & a & b \\0 & 1 &a\\0 &0 &1\end{bmatrix}$$ 
I tried inductive method to show $$A^n=\begin{bmatrix}1 & na & nb+\frac{n(n-1)}{2}a^2 \\0 & 1 &na\\0 &0 &1\end{bmatrix}$$ 
now : My question is : Is there other method (idea ) to find $A^n$ ?
Thanks in advance.
Can the idea apply for $$A=\begin{bmatrix}1 & a & b \\0 & 1 &c\\0 &0 &1\end{bmatrix}$$ when $c \neq a$ ?

$$A=I+aJ+bJ^2$$
where
$$J=\pmatrix{0&1&0\\0&0&1\\0&0&0}.$$
Then $J^3=O$.
Now
$$A^n=\sum_{k=0}^n{n\choose k}(aJ+bJ^2)^n
=I+n(aJ+bJ^2)+{n\choose 2}(aJ+bJ^2)^2$$
as $(aJ+bJ^2)^3=O$. But $(aJ+bJ^2)^2=a^2J^2$ so
$$A^n=I+naJ+nbJ^2+\frac{n(n-1)}2a^2J^2
=\pmatrix{1&na&nb+\frac{n(n-1)}{2}a^2\\0&1&na\\0&0&1}.$$
