$\newcommand{\eqvm}[1]{\underset{\small #1}{\equiv}}$
As already observed, $5777=53\times 109$. We can prove for divisibility of $P_T:=\displaystyle \prod_{i<j} (p_i^{p_j}-p_j^{p_i})$ by $53$ easily and then adapt the method for $109$.
If we have two primes $p$ and $q$ such that both $p \eqvm{53} q$ and $p \eqvm{52}q$, then it is clear from Fermat's Little Theorem that $p^q \eqvm{53} q^p$. If we discard primes $2, 13$ and $53$ (if among the given primes), all other primes are coprime to both $53$ and $52$, meaning a choice of $\phi(53)\phi(52)=52\cdot 24 = 1248$ possible different combination values of $\bmod (53,52)$, and we have at least $2014$ primes to consider, so by the pigeonhole principle we will have $p^q - q^p\eqvm{53} 0$ for some $p,q$ in the set (in fact for hundreds of cases).
If we attempt to use the same process for divisibility by $109$, we end up with $\phi(109)\phi(108)=$ $108\cdot 36 =3888$ pigeonholes; too many for our $2014$ pigeons (since $108=2^23^3$, we may be excluding $2,3$ and $109$ this time, if present).
But we can reduce the number of pigeonholes by considering the multiplicative order of each possible equivalence class $\bmod 109$. $1$ only has one slot available, since $1^j \equiv 1^k$ for any $j,k$. Also $108^j \equiv 108^k$ for any $j,k$ in our permitted primes, since they are all odd. The two order-$3$ values, $\{45,63\}$, create $2$ cases each and likewise the $2$ order-$4$ values $\{76,33\}$. The $2$ order-$6$ values $\{46,64\}$ also have two cases, since all exponents are $1$ or $5 \bmod 6$. The $6$ order-$9$ and $6$ order-$18$ have $6$ cases each, $4$ order-$12$ have $4$ cases, $18$ order-$27$ and $18$ order-$54$ have $18$ cases each, $12$ order-$36$ have $12$ cases and finally the $36$ order-$108$ primitive roots have $36$ cases each. This brings the total pigeonhole count to $1+1+4+4+36+36+16+324+324+144+1296=2190$.
Still too many pigeonholes. However let's look at the primitive root set (order-$109$) $\{6,10,11,13,$ $14,18,24,30,$ $37,39,40,42,$ $44,47,50,51,$ $52,53,56,57,$ $58,59,62,65,$ $67,69,70,72,$ $79,85,91,95,$ $96,98,99,103\}$ a little more and show that we can fill only (at most) half those pigeonholes. So far we have only considering that for $p \eqvm{(109,108)} (p_d,p_o)$ we cannot allow a $q \eqvm{(109,108)} (p_d,p_o)$. However if we consider some other $q \eqvm{(109,108)} (q_d,q_o)$ this may still be ineligible if $p_d^{q_o}
\eqvm{109} q_d^{p_o}$. And in fact because we are dealing with only primes which are coprime to $108$, every solution to $p_d^{q_o}$ is a primitive root too and for each $(p_d,p_o)$ there is always a corresponding valid $q_o$ for each of the $35$ other $q_d$ classes $\bmod 108$ in the primitive root set. So for example if we have a prime $p\eqvm{(109,108)} (11,5)$ then $\color{red}{11}^{43} \eqvm{109} 6^{\color{red}5}$, $\color{red}{11}^{103} \eqvm{109} 10^{\color{red}5}$, $\color{red}{11}^{73} \eqvm{109} 13^{\color{red}5}$, etc. So choosing $p\equiv (11,5)$ eliminates choices $(6,43), (10,103), (13,73)$, etc covering all other primitive roots$\bmod 109$. Thus if we chose all $36$ possibilities for $11 \bmod 109$ we would not be able to use any choices at all from other primitive root classes. The best we could hope for is to choose $18$ options for each primitive root which eliminates the other $18$ options for all other primitive roots, giving $18\cdot 36=648$ options and thus we now have more prime pigeons $(2014)$ than available pigeonholes (now $1542$), so $P_T$ is indeed divisible by $109$.
An "after-proof" thought:
"The best we could hope for" relies on the possibility that the exclusion relation between candidate primes is not transitive, but experimentally this is not the case. In the example for excluded cases from $(11,5)$ above, we find $13^{43} \eqvm{109} 6^{73}$, for example. In fact using $a'$ to denote the multiplicative inverse of $a\bmod 108$, we can see that
$p^{q} \eqvm{109} q^{p} \iff p^{p'} \eqvm{109} q^{q'}$. This ensures that exclusion is always transitive and we thus only actually have $36$ "pigeonholes" for the primitive root cases. Likewise using inverses to the multiplicative order of each class in turn, there are only $108$ pigeonholes overall.