Mathematical induction proof $n^4-1$ is divisible by $16$ for all odd integers $n$ I'm stuck towards the end of proving this, here's my attempt:
$P(3) = 80/16 = 5$, True
$P(k) = k^4-1$
$P(k+1)= (k+1)^4-1$
Expanded $= k^4+4k^3+6k^2+4k+1-1$
This is where I am stuck at.
Sorry for the sloppy formatting im still reading how to  properly write formulas on this site.
Thanks in advance for your help!

Without loss of generality we proved the statement for positive odd integers.
Note that a positive odd integer has the form of $2k-1$ where $k\ge 1$ 
We proceed with  induction on k.
Let $P(k)$ be  $(2k-1)^4-1$ is a multiple of $16$
For $k=1$, $(2k-1)^4-1=0$ which is a multiple of $16$ so $P(1)$ is true.
Assume that $(2k-1)^4-1$ is a multiple of $16$
We need to show that $(2k+1)^4-1$ is also a multiple of $16$
Note that the binomial theorem implies $$(2k+1)^4 = [(2k-1)+2]^4 = (2k-1)^4+8(2k-1)^3 +24(2k-1)^2+32(2k-1)$$
Thus $$(2k+1)^4 - (2k-1)^4 =8(2k-1)^3 +24(2k-1)^2+32(2k-1)=$$
$$ 8(2k-1)[(2k-1)^2+3(2k-1)+4)]$$
Note that the expression in square bracket is an even integer, so the result is multiple of $16$ 
