Note that $5777=53\cdot 109$. Partition $\left(\mathbb Z /p\mathbb Z\right)^{\times}$ into equivalence classes so that $x\sim y\iff p\mid x-y\text{ and }x,y\equiv k\cdot \text{ord}_p(x)\pmod{p-1}$. The rest should be PHP.

rafayaashary1 wrote: Note that $5777=53\cdot 109$. Partition $\left(\mathbb Z /p\mathbb Z\right)^{\times}$ into equivalence classes so that $x\sim y\iff p\mid x-y\text{ and }x,y\equiv k\cdot \text{ord}_p(x)\pmod{p-1}$. The rest should be PHP. Did you try to compute the bound you get? If I correctly understand what you're trying to do, I think this leads to a worse bound.

WLOG all primes are distinct. Throw $2,3,13,53,109$ if they exist. So we have at least $2000$ distinct primes. WLOG they are $p_1,p_2,\dots p_{2000}$. For each $p_i$, let $b_i=\frac{1}{p_i}\pmod {52}$. By PHP, there exist $i\neq j$ such that $p_i^{b_i}\equiv p_j^{b_j}\pmod {53}$. Hence by FLT $p_i^{p_j}\equiv p_i^{b_i p_i p_j}\equiv p_j^{b_j p_i p_j}\equiv p_j^{p_i}\pmod {53}$ Similarly we do it modulo $109$. Hence $\prod_{i<j} (p_i^{p_j}-p_j^{p_i})$ is divisible by $53\cdot 109=5777$