Prove p ⇒ (q ⇒ p) using the Fitch System No premises given.
In addition, any tips for doing Fitch problems? In particular, how should one begin them? The trickiest thing for me is figuring out exactly where to start. Once I get that figured out, the rest usually comes pretty quickly.
Thanks!
 A: $\def\fitch#1#2{~\begin{array}{|l}#1\\\hline#2\end{array}}$
Well now, $p\to(q\to p)$ effectively states: "If we first assume $p$, then if we subsequently assume $q$, we will find that $p$ is (already assumed) true."   Which is obvious; but this also tells us how the fitch proof is arranged: make two assumptions, restate the first, then make two deductions to discharge those assumptions.
$${\fitch{}{\fitch{p\hspace{15ex}\text{Assumption}}{\fitch{q\hspace{13.5ex}\text{Assumption}}{p\hspace{13.5ex}\text{Restatement}}\\ q\to p\hspace{10.5ex}\text{Deduction (conditional introduction)}}\\ p\to(q\to p)\hspace{5.5ex}\text{Deduction (conditional introduction)}}\\\blacksquare}$$ 
A: For Fitch proofs in general: typically your goal will give you the 'proof plan'.
In this case, for example, your goal is a conditional, so you'll want to set this up as a conditional proof, i.e. a $\to \: Intro$:


*

*$\qquad P$ Assumption (assumption of subproof, that is)


.
. (skip some lines)
.
n. $\quad Q \to P$  (desired last line of subproof ... we'll worry about how to get it later)
n+1. $P \to (Q \to P)$  $\to Intro$ 1-n
OK, so now we have a new goal: the $Q \to P$ that is the last line of the subproof.  Since that is a conditional itself, once again we will set this up with a conditional proof:


*

*$\qquad P$ Assumption (assumption of subproof, that is)

*$\qquad \qquad Q$ Assumption (assumption of subproof with subproof)
. (skip some lines)
.
n-1. $\qquad \qquad P$ (desired last line of inside subproof)
n. $\qquad Q \to P$  $\to Intro$ 2 - n-1
n+1. $P \to (Q \to P)$  $\to Intro$ 1-n
OK, so that's the plan .. now you just have to figure out how to get $P$ as the last line of the inside subproof, and you're there!
A: With only the two terms in the goal for the proof and no premises, you'll need to assume at least one term.
Maybe not the most elegant solution, but this worked for me:
1. | p ------------- assumption  
2. | | q ----------- assumption  
3. | | q & p ------- and introduction 1,2  
4. | | q ----------- and elimination 3  
5. | | p ----------- and elimination 3  
6. | q => p -------- implication introduction 2,5
7. | p => (q => p) - implication introduction 1,6  

Of note, 
1. | p ------------- assumption  
2. | | q ----------- assumption  

is insufficient to make an implication introduction of (p => q) but I am not entirely sure why...
