Added: As remarked in the answers below, my question has a negative (and well-known) answer.
We denote by $\mathcal P=\lbrace 2,3,5,7,\ldots\rbrace$ the set of prime-numbers and by $\mathcal P^*=\lbrace 2,3,4,5,7,8,9,1,13,16,\ldots\rbrace$ the set of non-trivial prime-powers. We consider the two functions $$A(n)=\prod_{p\in\mathcal P,p\leq n}p$$ (the product of all primes up to $n$, usually called the primorial function and written $\sharp n$) and $$B(n)=\prod_{q\in\mathcal P^*,q\leq n}q$$ (the product of all prime-powers up to $n$).
Is it true that $$A(n)<e^n<B(n)$$ for $n\geq 8$?
Short answer: No (both inequalities fail infinitely often, see Wojowu's answer below).
(This has certainly been studied but I am lousy at finding references.)
Remarks: The inequalities hold up to $10^6$.
The figure
gives values of $\sqrt{n}-\frac{1}{\sqrt{n}}\log(A(n))$ and $\sqrt{n}-\frac{1}{\sqrt{n}}\log(B(n))$ for all prime-powers up to $n=10^5$.
The normalization $\frac{n-\log(X(n))}{\sqrt{n}}$ (for $X\in\lbrace A,B\rbrace$) is suggested by the obvious identity $$B(n)=\prod_k A(\lfloor n^{1/k}\rfloor)^k$$ suggesting convergency of $$\frac{\log(B(n))-\log(A(n))}{\sqrt{n}}$$ (with values
at prime-powers up to $10^5$) to $2$.
Suitably strong versions of the prime-number theorem should give $$\lim_{n\rightarrow\infty} \frac{1}{n}\log(A(n))=\lim_{n\rightarrow\infty} \frac{1}{n}\log(B(n))=1.$$
D. Hanson (On the product of primes) gives the inequality $A(n)<3^n$ which is weaker. It seems that slightly stronger results are known but I could not find the inequalities above (which are perhaps not true!)
One can also consider the function $C(n)=\mathrm{lcm}(2,3,4,\ldots,n)$ defined as the product of all largest prime-powers $\leq n$. The number $C(n)$ is obviously a multiple of $A(n)$ and a divisor of $C(n)$. The function $C(n)$ is the exponential of second Chebyshev function.
It seems to oscillate (very irregularly) around $e^n$ as suggested by the graph
depicting the values of $\frac{n-\log(C(n))}{\sqrt{n}}$ at prime powers up to $10^5$.
