TITLE: $n\#$ (the $n$th primorial) seems very close to $e[(n-1)^n - n^{n-1}]$
QUESTION [5 upvotes]: Let $$n\# := P(n) P(n-1) \cdots P(2) P(1) ,$$ where $P(1)$ is the smallest prime, $2$, $P(2) = 3$, and so on. Then, as an example of what is claimed in the above title, $$P(11) = 200\,560\,490\,130, \qquad \textrm{and} \qquad 10^{11}-11^{10}=74\,062\,575\,399,$$ which is very close to $1/e$ times the first number. Does this continue to hold?

If so, I bet it is already known, but I’d like to know where to find it. There may be a connection with the Chebyshev function on Wikipedia, but that seems more akin to $e^{P(n)}$.

REPLY [5 votes]: To simplify your expression,
$\begin{array}\\
e[(n-1)^n - n^{n-1}]
&=en^{n-1}[(n-1)(1-1/n)^n - 1]\\
&=en^{n-1}[(n-1)(\frac1{e}-\frac1{2en}+O(\frac1{n^2}) - 1]\\
&=en^{n-1}[\frac{n}{e}-\frac1{2e}+O(\frac1{n})-\frac1{e}+O(\frac1{n}) - 1]\\
&=en^{n-1}[\frac{n}{e}-\frac{3}{2e}-1+O(\frac1{n})]\\
&=n^{n}-n^{n-1}(\frac32+e)+O(n^{n-2})\\
\end{array}
$
According to
https://en.wikipedia.org/wiki/Primorial,
$n\# = e^{n(1+o(1))}
$,
which does not match your expression.
However,
the standard definition of $n\#$
is the product 
of the primes up to $n$.
Your definition is
the product of the first $n$ primes.
Since the $n$-th prime
is about $n\ln(n)$,
this gives
$e^{(1+o(1))n\ln(n)}
=(e^{\ln(n)})^{(1+o(1))n}
=n^{(1+o(1))n}
$
and this does match your expression.
Note that
$n^n$ would work just as well,
since
$n^{n-1} = o(n^n)
$
and
$e(n-1)^n
=en^n(1-1/n)^n
=n^n(1+o(1))
$.