How to show $15 \mid 2^{4n}-1$ by induction 
For all $n \ge 1$ use mathematical induction to establish the divisibility of the statement: 
  $$
15 \mid 2^{4n}-1
$$

So first i substituted $1$ in and proved the statement to be true by example 
then for my assumption I put $k$ in for $n$ and for my proof so far I have:
\begin{align*}
15 &\mid 2^{4(k+1)}-1 \\
RHS & =2^{4k+4} -1 \\
&=2^4 \cdot 2^{4k}-1
\end{align*}
Is the splitting in the last step correct? Also how do I/where do I substitute my induction hypothesis into the equation? step by step explanation please!
 A: Your work is good so far.  Now notice that
$$
2^4 \cdot 2^{4k}-1 = 16 \cdot 2^{4k} - 1 = (2^{4k} - 1) + 15 (2^{4k})
$$
Your induction hypothesis is that $15 \mid 2^{4k }- 1$.  Also, it is true that $15 \mid 15(2^{4k})$.  So what do you know about the sum?
A: I propose to you an alternative proof. Note that $$2^{4n}-1=16^n-1$$
And $p(x)=x^n-1$ is divisible by $x-1$ by evaluating at $x=1$ for $p(1)=0$.
A: The statement is
$$
P(n):\quad 15 \mid 2^{4n} - 1.
$$
You have proved that $P(1)$ is true. Now you need to prove that if $P(k)$ is true then $P(k+1)$ is true. So you may assume that 
$$
15 \mid 2^{4k} -1.
$$
to prove that 
$$
15 \mid 2^{4(k+1)} -1
$$
is true.
But if you unpack this a bit
$$\begin{align}
2^{4k + 4} - 1 &= 2^4\cdot2^{4k} -1 \\&= 16\cdot 2^{4k} - 1 \\&= (15 + 1)\cdot 2^{4k} - 1\\ &= 15\cdot2^{4k} + 1\cdot 2^{4k} - 1 \\&=15\cdot2^{4k} + (2^{4k} - 1)
\end{align}
$$
and both of these terms are divisible by $15$ (using your hypothesis). That is, assuming that $P(k)$ is true, you have proved that $P(k+1)$ is true.
