Natural Deduction stuck $$\vdash (( p → q) →  p) →  p$$ how would I go about solving this with no premises. to get the implication i know i need $$(p → q) →  p$$ as my first assumption but im stuck to where to go from there
First hint: Prove ($\lnot$p$\rightarrow$(p$\rightarrow$q)) as a lemma.

More detailed analysis:


What is the goal?


The goal is to produce to (((p $\rightarrow$ q) $\rightarrow$ p) $\rightarrow$ p).


What evidence do you have so far?


You have evidence that if you assume ((p$\rightarrow$q)$\rightarrow$p) and produce p, that a solution can get constructed.


If you assume something, what is the structure of that assumption?


The assumption ((p$\rightarrow$q)$\rightarrow$p), has the structure such that (p$\rightarrow$q) is the antecedent and p is the consequent.  Thus, if you can produce (p$\rightarrow$q), by detachment ("modus ponens"), you can deduce p.


Can you assume anything and get rid of that assumption?


Suppose you assume ((p$\rightarrow$q)$\rightarrow$p).  You want 'p'.  But, let's say your enemy says that obtaining 'p' is impossible.  So, let's assume your enemy correct, and thus assume that $\lnot$p makes sense.  Can you then obtain 'p' under that assumption, with the analysis of 3. in mind?
I just did it with a truth table. Set p to each of 0 and 1 (false and true), and set q to each of 0 and 1 as well. That gives you 4 combinations of the two. Then, work your way from the inside out.


  Case 1: $p = 0$, $q = 0$


$p→q = 1$

$1→p=0$

$0→p=1$


  Case 2: $p = 0$, $q = 1$


$p→q = 1$

$1→p=0$

$0→p=1$


  Case 3: $p = 1$, $q = 0$


$p→q = 0$

$0→p=1$

$1→p=1$


  Case 4: $p = 1$, $q = 1$


$p→q = 1$

$1→p=1$

$1→p=1$

Therefore the statement holds true for every case.