Reason for LCM of all numbers from 1 .. n equals roughly $e^n$ I computed the LCM for all natural numbers from 1 up to a limit $n$ and plotted the result over $n$.  Due to the fast-raising numbers, I plotted the logarithm of the result and was surprised to find a (more or less) identity curve ($x=y$).
In other words, $LCM(1, 2, 3, ..., n)$ appears to be roughly the value $e^n$.
Is a there a simple explanation on why this is so?
$LCM(a, b, c, …)$ shall be defined as the least common multiple of all arguments $a, b, c, …$

 A: I think this is basically a different way to state exactly what André Nicolas said in a comment, but observe that the largest power of a given prime $p$ found as a factor in a number less than $n$ is explicitly given by
$$
\lfloor\log_p(n)\rfloor=\left\lfloor\frac{\ln n}{\ln p}\right\rfloor
$$
so we can write $LCM(n)=LCM(1,2,...,n)$ explicitly as
$$
LCM(n)=\prod_{p\leq n}p^{\left\lfloor\frac{\ln n}{\ln p}\right\rfloor}
$$
Applying $\ln(x)$ to both sides, we obtain
$$
\begin{align}
\ln(LCM(n))&=\sum_{p\leq n}\left\lfloor\frac{\ln n}{\ln p}\right\rfloor\cdot\ln p\\
&\approx \sum_{p\leq n}\frac{\ln n}{\ln p}\cdot\ln p\\
&=\ln n\cdot\sum_{p\leq n}1\\
&=\ln n\cdot \pi(n)
\end{align}
$$
where $\pi(n)$ denotes the function counting primes less than or equal to $n$. Now since by the prime number theorem we have $\pi(n)\sim\frac{n}{\ln n}$ so that these converge asymptotically, we get closer and closer to equality in the following approximation for $\ln(LCM(n))$:
$$
\begin{align}
\ln(LCM(n))&\approx\ln n\cdot\pi(n)\\
&\approx\ln n\cdot\frac{n}{\ln n}\\
&=n
\end{align}
$$
Applying the inverse of $\ln(x)$, namely $\text{e}^x$, then shows your statement. Note that the rounding off done by the floor functions becomes asymptotically insignificant.
