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 $16$.

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 $$(k^2 - 1)(k^2 + 1) = 8b(8b + 2) = 16b (4b + 1)$$
