Bogus Proof by Strong Induction So here is a bogus proof. 
Let $$P(n) ::= \forall k  \leq n, a^k =1, $$  
where k is a nonnegative valued variable. 
Base Case: $ P(0) $ is equivalent to $a^0 = 1$, which is true by definition of $a^n$ 
Inductive Step: By induction hypothesis, $a^k = 1 $ for all k $\in \mathbb{N}$ such that $k \leq n$. But then 
$$ a^{n+1} = \frac{a^n \times a^n}{a^{n-1}} = \frac{1 \times 1}{1} = 1$$which imples that $P(n+1)$ holds. It follows by induction that $ P(n)$ holds for all $n \in \mathbb{N}$. 
So I know this is bogus because we assumed that $P(n)$ is valid for $n \geq 1 $ even though this was not our base case. So we cannot assume that $ a^{n-1} = 1$ 
However, if I were to use the strong induction I am not quite sure why my logic fails. 
Base case is the same. 
Inductive Step: Assume that $P(0), P(1), ... P(n-1)$ is true. 
Then I just need to show that  $P(n) ::= \forall k  \leq n + 1,  a^k =1 $ is implied by my induction hypothesis. 
So for integers up to $n-1$,  $a^{n-1} = 1$ by inductive hypothesis. 
So I am only required to show that $a ^ {n} = 1$ 
$a ^ {n} = a^ {n-1} * a ^1 = 1 * 1 = 1$
Where did I fail?
 A: You know $P(1)$ is false. So see where the inductive step fails for $n=0$; how can you divide by $a^{n-1}$?
A: I know this in the following ironical form.
Theorem: Nobody understands induction.
Proof. Let $A(n)$ be the statement "In any group of $n$ people, nobody understands induction". Let's prove this by induction.
$A(0)$ is trivial.
Assume $A(n)$ is true.
Proof of $A(n+1)$: Consider any group of $n+1$ people. Removing one person $P$, we obtain a group of $n$ people; nobody of them will understand induction. Now pick one these $n$ people, say $P'$. Replacing $P'$ by $P$ we obtain another group of $n$ people and by $A(n)$ we conclude that $P$ (who now belongs to the group) does not understand induction. Thus none of the original $n+1$ people understands it.
To see where this "proof" fails consider $A(0+1)$.
A: The first proof fails because it makes $P(n+1)$ depend on $P(n-1)$ and $P(n)$ (since the calculation of $a_{n+1}$  makes reference to both $a_n$ and $a^{n-1}$), so this proof would need two base cases, rather than $1$.
Indeed, and much more simply put: you can't get $P(1)$ from $P(0)$ alone.
The second one (yours) does indeed not have that problem, but fails for a different reason:

$a ^ {n} = a^ {n-1} * a ^1 = 1 * 1 = 1$

No, that should be:
$a ^ {n} = a^ {n-1} * a ^1 = 1 * \color{red}a$
And note: you cannot save this proof by showing $a^1=1$, since all you are able to show is the base case $0$, i.e. that $a^0 = 1$
