Prove $5\mid2^{4n}-1$ by induction 
For all $n\ge1$, use mathematical induction to establish each other the following divisibility statements:
  $$5\mid2^{4n}-1$$ 

I was wondering if someone could help me with the set up of this proof. I only have done proofs with mathematical induction when I am given a series of numbers and I can incorporate my induction hypothesis in for one of my terms in the series. 
Step by step explanation please?
 A: Since you're new to induction, I'll give the framework and let you fill in the details. The induction step goes something like this:
Suppose that $5| 2^{4n} - 1$, and write $5k = 2^{4n} - 1$. Then
\begin{align*}
2^{4(n + 1)} - 1 &= 2^{4n} \cdot 2^4 - 1 \\
&= (5k + 1) \cdot 16 - 1 \\
&= 5k + 16 - 1
\end{align*}
Can you finish it from here, justifying each step?
A: Since you asked for a set-up,


*

*Show the statement is true for $n=1$, i.e.


$$5 \ \Big| \ 2^4-1$$


*

*Let $n\in\mathbb{N}$ be fixed. Assume for all $k<n$ that


$$5 \ \Big| \ 2^{4k}-1$$
Show that 
$$5 \ \Big| \ 2^{4n}-1$$
A: If $\displaystyle F(k)=2^{4k}-1,$
$\displaystyle F(k+1)-2^4\cdot F(k)=2^{4(k+1)}-1-2^4[2^{4k}-1]=2^4-1=15$
$\displaystyle\implies F(k+1)$ will be divisible by $15,$ (hence by $5$) if $F(k)$ is
Now establish the base case i.e., $k=0$
A: Hint $\ $ Let $\,f_n = 2^{4n}\!-1$. Prove the base case $\,5\mid f_0.\,$ Next prove $\,5\mid \color{#0a0}{f_{n+1}\!-f_n},\,$  therefore $\,5\mid \color{#c00}{f_n}\,\Rightarrow\, 5\mid \color{#0a0}{f_{n+1}\!-f_n}+\color{#c00}{f_n} = f_{n+1},\ $ i.e. $\ 5\mid f_n\,\Rightarrow\,5\mid f_{n+1} \,$  (the induction step).
Remark $\ $ Unwinding the above telescopic cancellation we can write $\,f_n\,$ as the sum of its differences $\ f_{k+1}-f_k = (16-1) 16^k$ which yields the following simple proof
$$ 5\mid 16-1\,\Rightarrow\, 5\mid 16^{n}-1\, =\, (16-1)\,(16^{n-1}+\cdots + 16^2 + 16 + 1)$$
See my posts on telescopy for many more examples
