Prove using mathematical induction: for $n \ge 1, 5^{2n} - 4^{2n}$ is divisible by $9$ I have to prove the following statement using mathematical induction.
For all integers, $n \ge 1, 5^{2n} - 4^{2n}$ is divisible by 9. 
I got the base case which is if $n = 1$ and when you plug it in to the equation above you get 9 and 9 is divisible by 9. 
Now the inductive step is where I'm stuck.
I got the inductive hypothesis which is $ 5^{2k} - 4^{2k}$
Now if P(k) is true than P(k+1) must be true. $ 5^{2(k+1)} - 4^{2(k+1)}$
These are the step I gotten so far until I get stuck:
$$ 5^{2k+2} - 4^{2k+2} $$
$$ = 5^{2k}\cdot 5^{2} - 4^{2k} \cdot 4{^2} $$
$$ = 5^{2k}\cdot 25 - 4^{2k} \cdot 16 $$
Now after this I have no idea what to do. Any help is appreciated. 

$\begin{align}{\bf Hint}\qquad\qquad\qquad\qquad\,\ \color{#c00}{25} &=\,\ \color{#c00}{16 + 9}\\
25^{\large N} &=\,\  16^{\large N}\! +\! 9j\\
\Rightarrow\,\ 25^{\large N+1}\! = \color{#c00}{25}\cdot 25^{\large N} &= (16^{\large N}\!+\!9j)(\color{#c00}{16+\!9}) = 16^{\large N+1} +9\,(\cdots)\ 
\end{align}$

Or, said mod $\,9\!:\,\ \begin{align} 25&\equiv 16\\ 25^{\large N}&\equiv 16^{\large N}\end{align}\ \Rightarrow\, 25^{\large N+1}\equiv 16^{\large N+1}\,$ by the  Congruence Product Rule
Or, $ $ equivalently, $\ \big[25\equiv 16\big]^{\large N}\!\Rightarrow\,  25^{\large N}\!\equiv 16^{\large N}\, $ by  the Congruence Power Rule, which is an inductive extension of the Product Rule.  
