Let be $p$ be a proposition. I want to show by means of a formal deduction that
$$((p \implies p) \implies p) \implies p$$
$$ ((p \implies p) \implies p) \implies p) \implies p) \implies p$$ $$ \vdots $$
$$ ((p \implies p) \implies p) \implies p) \cdots ) \implies p$$
for an odd number of implications distributed to the left, are all formal theorems. So far I have tried deducing
$$((p \implies p) \implies p) \implies p$$
But had no success...
Hint
By induction starting with the proof of the tautology : $p \to p$.
Then assume that it holds with $2n+1$ $\to$ signs and prove it for $2n+3$.
The idea is to assume the formula with $2n+2$ $\to$ signs. But by the induction hypo we know that the formula with $2n+1$ is a theorem; thus by Modus Ponens we derive $p$.
Then conclude using the Deduction Th.