what is the fault in this reasoning by strong induction
For all $ A $ and $ B $ of $ M_p (K) $ and all integer $n$ we have: $ A ^ n B = B $
The proof :
Denote $\forall n\in \mathbb N,\quad P (n) $ : $ A ^ n B = B $
The property is true at rank $ n = 0 $ because $ A ^ 0 B = I_n B = B $
Let $ n $ be a natural integer, suppose $ P (n) $ true up to rank $ n $. $ A ^ {n + 1} B = A A ^ n B = AB $ (hypothesis at rank $ n $) and $ A B = B $ (assumption at rank $ 1 $) we thus obtain $ A ^ {n + 1} B = B $
and $ P (n + 1) $ is true
In the inductive step you're only allowed to use previous steps. $P(1)$ is not a previous step for $P(1).$
The inductive step requires that $P(1)$ is already true, which is valid for $P(n)$ for all $n \geq2,$ but since you only have $P(0)$ as a base case you needed an inductive step that is valid for all $n \geq 1.$