Induction and contradiction I want to prove a statement by induction , I tasted the base case, then I considered the induction hypothesis for $n$ , so I assumed by absurdity that is not true for $n + 1$ but this contradicts the induction hypothesis , this proof is correct ? This idea is right ? If I suppose that is not valid for $n + 1$ and this contradicts the induction hypothesis , then this is proof?

Given a property $P$ and a base case, $n_0$, the following implication is verified (this is simple induction law):
$$ \left[P(n_0) \land \forall n \geq n_0\color{blue}{\Big(P(n) \implies P(n + 1)\Big)}\right] \implies \forall n \geq n_0 \Big(P(n)\Big)$$
If we change the blue part by another equivalent expression, then the main implication will still being correct (so you will be able to argue as described by the result expression).
Consider logical variables $p$ and $q$ defined as:
$$p = P(n) \text{ is verified}$$
$$q = P(n + 1) \text{ is verified}$$
So, by logical equivalences:
$$\begin{align}p \to q &\equiv \lnot  p \lor q\\
&\equiv\lnot p \lor q \lor \text{false}\\
&\equiv\lnot (\lnot(\lnot p \lor q)) \lor \text{false}\\
&\equiv\lnot ( p \land \lnot q) \lor \text{false}\\
&\equiv(p \land \lnot q) \to \text{false}
\end{align}$$
That is exactly what you refere to: considere as a hypothesis [$P(n)$ true and $P(n+1)$ false] and get a contradiction.
So, your method works:
$$ \left[P(n_0) \land \forall n \geq n_0\color{blue}{\Big(P(n) \land \lnot P(n+1)\implies \text{false}\Big)}\right] \implies \forall n \geq n_0 \Big(P(n)\Big)$$
