I was playing around with numbers and I discovered that $$\left\lfloor e^{\frac{p_4\#}{p_5}}\right\rfloor=\left\lfloor e^{\frac{210}{11}}\right\rfloor =13981^2,$$ with floor function $\lfloor x \rfloor := \max\{m\in\mathbb{Z} : m\leqslant x\}$; $ \ p_n$ denotes the $n^{\text{th}}$ prime number; $ \ $and primorial $$p_n\#_c := \prod_{i=1}^n (p_i + c).\tag{$p_n\#_0 = p_n\#$}$$
Furthermore, $$\left\lfloor e^{\frac{p_1\#}{p_2}}\right\rfloor = 1^2.$$ I then made a conjecture with very few support, that $$\left\lfloor e^{\frac{p_{n^2}\#}{p_{n^2 + 1}}}\right\rfloor \tag1$$ is always a square number, say $k_n^{\ \ 2}$.
Does somebody have a big enough computer to find the value of $(k_n)_{n\geqslant3}$? Or can my conjecture be proven/disproven with a pen and paper? And perhaps, to support the potential of this not being a coincidence, every divisor of $13981$ takes the form $(13981 - 10x)$ for some $x\geqslant 0$. Maybe if this conjecture is true, $k_n > 1$ has this certain property?
Edit 1: With some computational power, I found that $k_3^{\ \ 2}$ has $\approx 176,000$ digits. It turned out, however, that I was wrong, and really, $k_3^{ \ \ 2}$ has $\approx 3,340,970$ digits.
Edit 2: Expressions equivalent to $(1)$, avoiding $e$, for any base $b\in\mathbb{N}_{>1}$:$$\left\lfloor b^{\frac{p_{n^2}\#}{p_{n^2 + 1} \cdot Ln(b)}}\right\rfloor=k_n^{ \ \ 2}$$E.g. base $b=2$:$$\left\lfloor 2^{\frac{p_{n^2}\#}{p_{n^2 + 1} \cdot Ln(2)}}\right\rfloor=k_n^{ \ \ 2}$$E.g. base $b=n^2$:$$\left\lfloor (n^2)^{\frac{p_{n^2}\#}{p_{n^2 + 1} \cdot Ln(n^2)}}\right\rfloor=k_n^{ \ \ 2}$$