How would one prove ((p ⇒ q) ⇒ p) ⇒ p, using the Fitch system

Question

FYI, the logic program I use cannot do contradiction introductions. This point is most likely irrelevant, for I highly doubt I would need to use any form of contradiction for this proof.

In my attempt to solve this, I started off with assuming (p ⇒ q) ⇒ p)
Is this correct?

If so, what next? Forgive me if the solution seems so obvious.

Answer 1

(p ⇒ q) ⇒ p
((p ⇒ q) ⇒ p) ∨ (p ⇒ p)        ; (X ⇒ X) and Or introduction
((p ⇒ q) ∨ p) ⇒ p              ; (X ⇒ Z) ∨ (Y ⇒ Z) |- (X ∨ Y ⇒ Z)
((¬p ∨ q) ∨ p) ⇒ p             ; (p ⇒ q) ⇔ (¬p ∨ q)
((¬p ∨ p) ∨ q) ⇒ p             ; (X ∨ Y) ∨ Z |- (X ∨ Z) ∨ Y
(true ∨ q) ⇒ p                 ; (¬X ∨ X) ⇔ true
true ⇒ p                       ; (true ∨ X) ⇔ true
p                              ; Implication elimination
((p ⇒ q) ⇒ p) ⇒ p              ; Implication introduction