Detect Wrong Proof by strong induction $a^{n-1}=1$ for all $n$
Obviously there is some error in the steps, but i can't figure it out, appreciate some hint.
We show using strong induction that $$ a^{n-1}=1 \hspace{1cm} \forall n\in \mathbb{N}$$
$$ n=1 \hspace{.5cm} \Longrightarrow \hspace{.5cm} a^{1-1}=1$$
We assume that is true for k with $1\leq k< n-1$, then $a^{n-1}=\frac{a^{n-2}a^{n-2}}{a^{n-3}}=\frac{1.1}{1}=1$
Therefore by strong induction principle. $\hspace{1cm}a^{n-1}=1\hspace{1cm} \forall n \in \mathbb{N}$
thanks
You have assumed that $n-3\ge 1$ or $n\ge4$. So you need to check for $n=2$ and $n=3$. Clearly, $a^{2-1}=a\neq 1$.