Find the smallest possible minimum difference for an arithmetic progression that contains 4 primes. How about 5 or 6? The solution makes sense, but I don't get how this ends up being true. So the solution is if $n$ is the number of primes you need. You get the product of all the primes between $[0,n]$ then that ends up being the minimum difference.
$n=4$
The solution is all primes < 4 are 2 and 3 so the minimum difference is 6 so the set becomes {5,11,17,23}
$n=5$
Solution is all primes < 5 so 2 and 3 again with a minimum difference of 6 so the set becomes {5,11,17,23,29}
Then $n=6$ ends up being a minimum difference of 30 because 2, 3 and 5 are the primes below 6.  
Why does this work? What's the theory behind it?
 A: The basic reason you generally need to use a difference which is a multiple of all primes less than $n$ is due to, otherwise, at least one of the values in any arithmetic progression being a multiple of a prime factor less than the # of primes you're trying to get. However, there's an exception to consider which I describe later.
For example, with some $n$, consider the difference $d$ not being a multiple of some prime $p \lt n$, say $n \equiv r_1 \pmod p$, with $0 \lt r_1 \lt p$. With the first prime in any sequence you're checking, $p_0$, have that $p_0 \equiv r_0 \pmod p$ (including the possibility that it could be $0$ if you're checking $p$ itself). Then the $i$'th $0$-based arithmetic progression term, $a_i$ would be $a_i \equiv r_0 + ir_1 \pmod p$. However, consider $0 \le i,j \lt p$ and $j \gt i$. Then $a_j - a_i \equiv (r_0 + jr_1) - (r_0 + ir_1) \equiv (j-i)r_1 \not\equiv 0 \pmod p$. This means each of these $p$ integers have distinct values modulo $p$, so they each must represent the values of $0$ to $p - 1$ in some order. If the first prime was actually $p$, so $r_0 = 0$, then the $p + 1$'st integer would be a multiple of $p$ and, thus, not a prime. Otherwise, one of the other first $p$ integers in the sequence is a multiple of $p$ and, thus, not a prime. In either case, this would mean the sequence cannot contain only primes.
However, one small exception to the above argument occurs if $n$ itself is prime, then unless there's a sequence of primes which starts at $n$ itself, then one of the later sequence terms will be a multiple of $n$ if the difference of $d$ is not a multiple of $n$. This is why $n = 3$ works since {$3,5,7$} is an arithmetic progression of $3$ primes which are each $2$ apart, and your example of $5$ also works. However, $n = 7$ doesn't work as among the $7$ integers {$7,37,67,97,127,157,187$}, the last one is not prime since $187 = 11 \times 17$. I haven't checked larger primes, but I suspect very few, if any, will work with $d$ being just a product of the primes smaller than $n$.
FYI, only just recently in $2004$, Ben Green and Terence Tao proved the Green–Tao theorem which states there are arithmetic progressions of only primes of length $k$ for every natural number $k$. The Wikipedia article explains some asymptotic formulas for the number of $k$-tuple of primes less than or equal to $N$. Also, there's some interesting numerical results of the first set of $24$, $25$ and $26$ primes in an arithmetic progression, with the $d$ for each one being $23$ primorial, i.e., the product of all of the primes less than or equal to $23$.
