**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

[![enter image description here][1]][1]

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

[![enter image description here][2]][2]

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

[![enter image description here][3]][3]

depicting the values of $\frac{n-\log(C(n))}{\sqrt{n}}$
at prime powers up to $10^5$.


  [1]: https://i.sstatic.net/7WMjo.gif
  [2]: https://i.sstatic.net/kHXD3.gif
  [3]: https://i.sstatic.net/aPxY6.gif