Identify the fallacy in strong induction proof I was wondering what the fallacy was in the following proof:

Given $a \in \mathbb{R}$, one has that $a^n=1$ for all $n \in \mathbb{N}\cup\{0\}$.
"Proof"
If $n=0$ then $a^0=1$. Thus the claim is true for the base case. Next, fix $n \in \mathbb{N} \cup \{0\}$ and assume that the claim is true for all $k\in \mathbb{N}\cup \{0\}$ with $k \leq n$, i.e., assume that $a^k = 1$ for all $n \in \mathbb{N}\cup \{0\}$ with $k \leq n$. We want to show that the claim is true for $n+1$. Observe that
$$a^{n+1} = \dfrac{a^n \times a^n}{a^{n-1}} = \dfrac{1 \times 1}{1} =1$$
where we have used the induction hypothesis in the second equality. Thus the claim is true for $n+1$ and by PMI we can now conclude that the claim is true for all $\mathbb{N} \cup \{0\}$. $\Box$

Clearly the claim is false, and I have to find the error in the proof, not just present a counterexample. So, I think that the error is that you cannot extend the induction hypothesis to any given $a \in \mathbb{R}$ since, by the Peano Property, the induction step only holds for any given $n \in \mathbb{N}$.
Is my reasoning correct? Please let me know.
 A: Your inductive step requires the assumption that the result holds (in particular) for $n$ and $n-1$. However, your base case only covers $n=0$ whereas you'd need $n=1$ also for this inductive step to apply.
Here's another example.
Claim  that everyone in a group of any countable size has the same age.
Base case: 
Suppose we have a set of 1 person. Then obviously everyone has the same age.
Now suppose the hypothesis is true for a set of $n$ people, say. $\{a_1, \cdots , a_n\}$.
Take a group of $n+1$ people, $\{a_1, \cdots , a_{n+1}\}$.
We can split this into two smaller subsets $\{a_1, \cdots , a_n\}$ and $\{a_2, \cdots , a_{n+1}\}$ in which each set of people have the same age, as each set has size $n$. Therefore $a_{n+1}$ has the same age as some other $a_i:1<i<n+1$ has the same age as $a_1$, so everyone in that set of $n+1$ people have the same age.
The issue with this proof is that we are using the fact that if $\{a_1, \cdots , a_n\}$ have the same age, any two of them do too. But this assumes that any group of two people $\{a_1, a_2\}$ have the same age which is not true in general.
For this to work you'd need to show the base case of $n=2$, but you cannot do this.
Moral of the story - be careful with your base cases and with what you're assuming when using induction!
