Induction: confused by $P(k)$ to $P(k+1)$ As far as I know, I know we have to use $P(k)$ to prove $P(k+1)$ in induction instead of going from $P(k+1)$ to $P(k)$.
I want to prove a divisibility problem using the idea of induction, and I can't show the problem here directly but I found a very similar one.
The example I saw online is this: Prove $6^n + 4$ is divisible by $5$ using induction. The URL to this problem is here: https://www.iitutor.com/mathematical-induction-divisibility/
I read his proof steps, I found it very confusing that he actually started the proof from $P(k+1)$, as he proceeded the proof, he confronted the situation where the induction hypotheses can help, then he applies the induction hypothesis to complete the proof.
But wouldn't this be not allowed? Since he actaully started from $P(k+1)$ instead of $P(k)$?
P.S. I have read and confronted several examples that started from $P(k+1)$ instead of $P(k)$, but our professor told us that is actually not logical correct. Which way is truly the correct way to do an induction? I am already lost...
 A: We want to prove that $6^n+4$ is divisible by $5$ for all natural numbers $n$. This proceeds in three steps.
$1.)$ We show that the statement $P(n)$, $6^n+4$ is divisible by $5$, is true for $n=1.$ In other words, we confirm that $P(1)$ is true.
$2.)$ One assumes that the statement $P(n)$ is true for some $k>1$. In other words we assume that $P(k)$ is true, that is, $6^k+4$ is divisible by $5$.
$3.)$ One uses the assumption made in $(2)$ to prove that $P(k+1)$ is true. In other words, we show that $6^{k+1}+4$ is divisible by $5$.
This process proves that $P(n)$ is true $\forall n\in \mathbb{N}$.
Theorem:
For all $n\in \mathbb{N}$, $6^n+4$ is divisible by $5$.
Proof:
$1.)$ Clearly $P(1)$ is true since $6^1+4=10$ is divisible by $5$.
$2.)$ Now assume that $P(k)$ is true for some $k>1$. That is, assume that $6^k+4$ is divisible by $5$.
$3.)$ Note that $$6^{k+1}+4=6\cdot6^k+4+20-20=6(6^k+4)-20$$ To show that $P(k+1)$ is true, we show that $$\frac{6^{k+1}+4}{5}=\frac{6(6^k+4)}{5}-\frac{20}{5}$$ is an integer. Well, since we assumed that $6^k+4$ is divisible by $5$, clearly $\frac{6(6^k+4)}{5}$ is an integer, as is $\frac{20}{5}=4$.
Therefore we conclude that $P(k+1)$ is true, that is we confirmed that $6^{k+1}+4$ is divisible by $5$.
A: No, he did not start from $P(k+1)$. He is simply stating what needs to be done in the beginning of Step $3$. That is, he is stating his goal in the first $2$ lines of Step $3$, and then proceeds to achieve that goal.
