How to proof $\forall n>0: 5|(6^n-1)$ is true using mathematical induction This is what i did so far, i got stuck on the later steps of using mathematical induction
I started with a base of $1$ which gave me $5|5$ which is $1$. Then I used $n = k$, and i got $5|(6^k-1)$ and i assumed this was true, but when i got to $n = k+1$ i didnt knew how to implement $5|(6^k-1)$ to $5|(6^{k+1}-1)$ to show that the theorem is true. For the real assignment, check this link: https://imgur.com/e6Z4tZW
 A: Hint:
$$5|(6^k-1)$$
implies
$$6^k=5p+1$$
for some integer $p$.
Or, write
$$6^{k+1}-1=6\times 6^k -1$$
$$=6*(6^k-1)+6-1$$
A: let $$T_n=6^n-1$$ and $$5|6^n-1$$ now let $$T_{n+1}=6^{n+1}-1$$ and we get
$$T_{n+1}-T_n=5\cdot 6 ^n$$ therefore $$5|T_{n+1}$$
A: You have assumed that $$5|(6^k-1)$$ This means that $6^k-1=5m$ for some (natural) number $m$. 
Firstly, to turn the $6^k$ into a $6^{k+1}$, multiply the whole equation by $6$ to give $$6^{k+1}-6=30m$$Then to get the $-1$, just subtract $1$ from both sides and rearrange a bit to give $$6^{k+1}-6-1=30m-1\\6^{k+1}-1=30m-1+6\\6^{k+1}-1=30m+5\\6^{k+1}-1=5(6m+1)$$
A: With induction. If $5|6^n-1$, then there is $q_n$ such that $6^n-1=5\cdot q_n$. Note that
$$
6^{n+1}-1=6\cdot 6^n-1=5\cdot 6^n+6^n-1=5(6^n+q_n)
$$
Without induction.
\begin{align}
6^n-1 
=&
(5+1)^n-1
\\
=&
(5^{n}+n\cdot 5^{n-1}+\frac{n(n-1)}{2}\cdot 5^{n-2}+\cdots+\frac{n(n-1)}{2}\cdot 5^2+ n\cdot5+1)-1
\\
=&
5^{n}+n\cdot 5^{n-1}+\frac{n(n-1)}{2}\cdot 5^{n-2}+\cdots+\frac{n(n-1)}{2}\cdot 5^2+ n\cdot5
\\
=&
5\cdot (5^{n-1}+n\cdot 5^{n-2}+\frac{n(n-1)}{2}\cdot 5^{n-3}+\cdots+\frac{n(n-1)}{2}\cdot 5+ n)
\\
\end{align}
A: Step: $5|(6^{n+1} -1)$
$6^{n+1} -1 = 6^n6 -1=$
$6^n (1+5) -1 = 6^n-1 +5(6^n).$
$5|(6^n -1)$  $\rightarrow$  $(6^n -1)= 5s$.
$6^{n+1} - 1 = 5s + 5(6^n) = 5(s+6^n).$
Hence:
$5|(6^{n+1} -1).$
