I do not have expert experience in mathematics so I expect the answer to be simple. I'm currently looking into mathematical induction and saw this problem (Question $2$):
Prove that for n any positive integer: $$6^n −1$$ is divisible by $5$.
The solution shows:
$$6^{k+1} − 1$$ $$= 6(6^k) − 1$$ $$= 6(6^k − 1) − 1 + 6$$ $$= 6(6^k − 1) + 5$$
My question is how does: $$6(6^k) − 1$$ equate to: $$6(6^k − 1) − 1 + 6$$
One way to see is that when you expand the final expression you have: $$6(6^k-1)-1+6=6(6^k)-6-1+6=6(6^k)-1$$ Where does this come from? This is a common thing in mathematics, whenever required, we add a "zero". See: $$6(6^k)-1=6(6^k)+(\color{blue}{-6+6})-1$$ You factor the $6$. (Leave a comment if you don't know what that means.) $$\implies 6(6^k)-1= 6(6^k-1)+6-1$$ Hope this helps. Ask anything if not clear :)
For the $\textbf{induction step},$ you prove that if the statement holds for any given $k$, then it must also hold for the case $k+1$. This is done using the $\textbf{induction hypothesis}.$
Step $1$: Base case. The statement holds for $n=1,2$
Step $2$: Inductive step. Assume that it holds for any arbitrary positive integer $k$, and we show it also holds for $k+1$. Then we use the induction hypothesis, that $6^{k}-1$ is divisible by $5$, to show that $6^{k+1}-1$ is also divisible by $5$. This is why you need to somehow introduce $6^{k}-1$.
$$6^{k+1} − 1$$ $$= 6(6^k) − 1 \quad\textbf{(1)}$$ $$= 6(6^k − 1) − 1 + 6\quad\textbf{(2)}$$ $$= 6(6^k − 1) + 5 \quad\textbf{(3)}$$
In step $\textbf{(1)},$ we have $6^{k}$, but need $6^{k}-1$, so we write $6(6^{k}-1),$ but this gives an extra $-6$. So to remove this error we add $6$ back. Thus we arrive at step $\textbf{(2)},$ which simplifies to the next step, where we then use that induction hypothesis to conclude that $6^{k+1}-1$ is also divisible by $5$.
Since $n=k$ is arbitrary, we conclude that $6^{n}-1$ is divisible by $5$ for any positive integer.