If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$? All is in the title:

If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$?

I can't conclude from the fact that $k^2 - 1$  is divisible by $8$, that then $k^4-1$ is divisible by $16$.
 A: Hint: $$k^4 - 1 = (k^2 - 1)(k^2 + 1) = (k^2 - 1)\Big((k^2 - 1) + 2\Big)$$

ADDED per comment: So yes, we have that if $(k^2 - 1)$ is divisible by $8$, then $$k^4 - 1 = (k^2 - 1)(k^2 + 1) = 8b(k^2 + 1)$$ for some integer $b$. 
And now, if $k^2 - 1$ is divisible by 8, it is even, then so is $k^2 + 1$. 
That is, $k^2 + 1 = (k^2 - 1) + 2 = 8b + 2$. So $$(k^2 - 1)(k^2 + 1) = 8b(8b + 2) = 16b (4b + 1)$$
A: $k^4 - 1$ can be factorised as $(k^2-1)(k^2+1)$ if the first term is divisible by $8$ means it's even, and so the term plus $2$ is also even,
And so the first term is of the type $8c$ and the next of type $2d$ finally it is $16$ times some number. 
A: $\displaystyle (2a+1)^2=4a^2+4a+1=8\frac{a(a+1)}2+1=8b+1$ where $b$ is some integer
Now, $(2a+1)^4=(8b+1)^2=64b^2+16b+1\equiv1\pmod{16}$
This way we can show $(2a+1)^{2^n}\equiv1\pmod{2^{n+2}}$ for integer $n\ge1$
This is generalized as  Carmichael function $$\lambda(2^{n+2})=2^n\text{ for }n\ge1$$
A: But much, much more than this is true. Indeed:
If $z=1+p^ru$, then $z^p=1+p^{r+1}u'$, as one sees easily from the binomial expansion. And even more is true, namely that, except in the case $p^r=2$, if $p$ doesn’t divide $u$, it doesn’t divide $u'$ either.
