Let $p_1, p_2, \dots, p_n$ be the first $n$ primes and $p_n\#$ be the primorial of $p_n$
Using the Chinese Remainder Theorem and Fermat's Little Theorem, the following, I believe, is true:
$$\sum_{i}^{n}\left(\frac{p_n\#}{p_i}\right)^{p_i-1} \equiv 1 \pmod {p_n\#} $$
Here's my reasoning:
Since $x^{p_i-1} \equiv 1 \pmod {p_i}$, it follows that the $\left(\frac{p_n\#}{p_i}\right)^{p_i-2}$ is the inverse of $\frac{p_n\#}{p_i}$ modulo $p_i$
So, it follows that there exists an $x$ such that:
$$\sum_{i}^{n}\left(\frac{p_n\#}{p_i}\right)^{p_i-1} = x{p_n\#} + 1$$
Is there a straight forward way to establish this expression without appeal to the Chinese Remainder Theorem or Fermat's Little Theorem. Does $x$ have any interesting properties?