Troubles with proving the following expression $((p \implies q) \implies p) \implies p$ using Fitch System I having heavy difficulties with this exercise beacuse it doesn't have any premise. I can't get the result I want (which is posted on the title of the question) straight to the top level
1.              p Assumption
2.                  (p => q) => p Assumption
3.                  p Reiteration: 1
4.              ((p => q) => p) => p Implication Introduction: 2, 3
5.          p => ((p => q) => p) => p Implication Introduction: 1, 4

ps: I don't want the entire proof of the expression. I would appreciate hints. If it is really complicated, then I accept the entire proof so I can study further.
ps: I know you all want to help, but I have only these operations available:
-assumption
-reiteration
-delete
-negation introduction
-negation elimination
-and introduction
-and elimination
-or introduction
-or elimination
-implication introduction
-implication elimination
-biconditional introduction
-biconditional elimination
It was cool to know Peirce's law, however I didn't find a way to implement it through the resources above.
 A: HINT
whenever you want to prove something of the form $\phi \rightarrow \psi$, do a conditional proof where you assume $\phi$ and try to get to $\psi$.
So in your case, assume $(p \rightarrow q) \rightarrow p$, and try to get to $p$
You did do a conditional proof, but notice that the conditional you obtained was not the one you want.
HINT 2
To get to $p$ inside the conditional proof ... Do a proof by contradiction
HINT 3
If you are still stuck ... See the 9th post under 'Related' on the right.
A: For a proof of Peirce's law with Natural Deduction's based Fitch system, you can see the answer to the post: using the Fitch system how do I prove $((p \implies q) \implies p) \implies p$ ?
If you are not falimiar with the rules using $\bot$ (the falsum) you can rewrite it as follows:
1) assume $(P \to Q) \to P$
2) assume [a]: $P$ and [b]: $¬P$ and derive by $¬$E: $Q$
3) derive $P \to Q$ by $→$I, discharging [a]
4) use it with 1) to derive $P$ by $→$E.
Now we have again a contradiction with the [b] assumption: $¬P$.
The next step needs $¬$I, to derive from the said contradicition: $¬¬P$, discharging [b].
Up to now, having discharged the "temporary" assumptions [a] and [b], we have:

$(P \to Q) \to P \vdash ¬¬P$.

The following step needs $¬¬$I (Double Negation elimination) to derive $P$ from $¬¬P$ and conclude with: $(P \to Q) \to P \vdash P$.
The final step needs $\to$I to conclude with:


$\vdash ((P \to Q) \to P) \to P$.


