Prove that $n^4-n^2$ is divisible by $8$ if $n$ is an odd positive integer. 
Prove that $n^4-n^2$ is divisible by $8$ if $n$ is an odd positive integer.

I'm supposed to use proof by induction, but I failed at it miserably. So far I have this:
$$(n^4) - (n^2) = (n^2)((n^2)-1) = n(n-1)n(n+1)$$
Let $n = 2k + 1$ because its always odd. Then
$$\begin{align}
&(2k+1)(2k)(2k+1)(2k+2)
\\\implies\quad &(2k+1)(2k)(2k+1)2(k+1)
\\\implies\quad &(2k+1)2(k)(2k+1)2(k+1)
\\\implies\quad &4(2k+1)(k)(2k+1)(k+1)
\\\implies\quad &4((2k+1)^2)(k)(k+1)
\\\implies\quad &4((2k+1)^2)((k^2)+k)
\end{align}$$
Now $(2k+1)^2$ is odd and $k^2 + k$ is always even because
$$\begin{cases}
\mathrm{odd}^2 + \mathrm{odd} = \mathrm{even} \\
\mathrm{even}^2 + \mathrm{even} = \mathrm{even}
\end{cases}$$
So that means I can take out a $2$ and write $((k^2)+k) = 2X$.
$$\begin{align}
&4(2X)((2k+1)^2)
\\\implies\quad &8X((2k+1)^2)
\end{align}$$
and because there is an $8$ I can say its always divisible by $8$, no matter what odd integer $k$ is? How can this be done using proof by induction?

Base step : $8$ divides $0$.
Hypothesis Step : Let $(2k+1)^4 - (2k+1)^2 = 8m$
Induction Step : To show : $(2k+ 3)^4 - (2k+3)^2 = 8j$
Now, the main insight is that you must somehow manipulate the induction step to use the hypothesis. How do we do that ?
$$
\begin{align}
&&(2k+1+2)^4-(2k+1+2)^2 \\
&&=(2k+1)^4 +4(2k+1)^3\times 2 + 6(2k+1)^22^2+4(2k+1)2^3 +2^4 - (2k+1)^2 -2(2k+1)2 -4 \\
&&= (2k+1)^4 - (2k+1)^2 + 8(2k+1)^3 + 8(2k+1)^23 + 8(2k+1)2^3 + 2^3\times 2 - 4(2k+1+1)\\
&&=8m + 8a - 8(k+1)\\
&&=8j
\end{align}
$$
Hence, proved.
The main insights of this proof are ...
1. We try to prove this on $k $ rather than $n$because it allows us to apply induction easier.
2.We write $3 $ as $1 +2$ so that we can use our hypothesis and build on what has already been done.
Simpler alternate proof : $A = n^2(n+1)(n-1)$
$n+1$ and $n-1$ are two consecutive even integers. So, one of them has to be a multiple of 4. This means that $A$ is divisible by 8.
