We begin with the equation: $e^{5(\log_e x)^2 + 3} = x^8, \; x > 0.$ Equating the exponents, we have: $5(\log_e x)^2 + 3 = \log_e x^8 = 8 \log_e x.$ Let $t = \log_e x$. Substituting, the equation becomes: $5t^2 + 3 = 8t.$ Rewriting this as a quadratic equation: $5t^2 - 8t + 3 = 0.$ Factoring the quadratic: $5t^2 - 5t - 3t + 3 = 0,$ $(5t - 3)(t - 1) = 0.$ Thus, the solutions for $t$ are: $t = 1$ implies $\log_e x = 1$, giving $x = e$. $t = \frac{3}{5}$ implies $\log_e x = \frac{3}{5}$, giving $x = e^{\frac{3}{5}}$. The product of all solutions is: $e^1 \times e^{\frac{3}{5}} = e^{1 + \frac{3}{5}} = e^{\frac{8}{5}}.$