<s>[INST] Is $\ 7!=5040\ $ the largest highly composite factorial? A highly composite number is a natural number $\ n\ge 1\ $ that has more divisors than any smaller natural number $\ m\ge 1$.
Checking the $\ 1000\ $ entries in the OEIS-sequence, I noticed that only the following factorials are highly composite :
$$[1, 2, 6, 24, 120, 720, 5040]$$

Is it known whether $\ 7!=5040\ $ is the largest highly composite factorial ?

The factorials have some conditions necessary for a number to be highly composite :


*

*They contain the first $k$ prime numbers as prime factors.

*The exponents are non-increasing.

*The exponent corresponding to the largest prime factor is $\ 1$.


So, it is not obvious whether the above list is complete.
 [/INST] Suppose $n \ge 20$. Trivially, $n!$ is then divisible by $16$, so $\dfrac{13}{16}n!$ is an integer smaller than $n!$.
Let the prime factorization of $n!$ be $n! = 2^{e_1}3^{e_2}\cdots11^{e_5}13^{e_6}17^{e_7}\cdots p_r^{e_r}$. 
Then the prime factorization of $\dfrac{13}{16}n!$ is $\dfrac{13}{16}n! = 2^{e_1-4}3^{e_2} \cdots11^{e_5}13^{e_6+1}17^{e_6}\cdots p_r^{e_r}$.
Then, the number of divisors of $n!$ is $\sigma_0(n!) = (e_1+1)(e_6+1)\displaystyle\prod_{k \neq 1,6}(e_k+1)$ and the number of divisors of $\dfrac{13}{16}n!$ is $\sigma_0(\dfrac{13}{16}n!) = (e_1-3)(e_6+2)\displaystyle\prod_{k \neq 1,6}(e_k+1)$. 
Hence, $\sigma_0(\dfrac{13}{16}n!) > \sigma_0(n!)$ iff $(e_1-3)(e_6+2) > (e_1+1)(e_6+1)$ iff $e_1 > 4e_6+7$. 
Using the well known formula for the largest power of a prime dividing a factorial, we have 
$e_1 = \displaystyle\sum_{k = 1}^{\left\lfloor \log_2 n\right\rfloor}\left\lfloor \dfrac{n}{2^k} \right\rfloor  > \displaystyle\sum_{k = 1}^{\left\lfloor \log_2 n\right\rfloor} \left(\dfrac{n}{2^k} - 1\right) \ge n - \dfrac{n}{2^{\left\lfloor \log_2 n\right\rfloor}} - \left\lfloor \log_2 n\right\rfloor \ge n - 2 - \log_2 n$.
Similarly, $e_6 = \displaystyle\sum_{k = 1}^{\left\lfloor \log_{13} n\right\rfloor}\left\lfloor \dfrac{n}{13^k} \right\rfloor \le \displaystyle\sum_{k = 1}^{\left\lfloor \log_{13} n\right\rfloor}\dfrac{n}{13^k}  \le \dfrac{n}{12}$. Hence, $4e_6+7 \le \dfrac{n}{3}+7$. 
I'll leave it as an exercise to show that $\dfrac{n}{3}+7 < n - 2 - \log_2 n$ holds for all integers $n \ge 20$. 
Therefore, for all integers $n \ge 20$, we have $4e_6+7 \le \dfrac{n}{3}+7 < n - 2 - \log_2 n < e_1$, and thus, $\sigma_0(\dfrac{13}{16}n!) > \sigma_0(n!)$. 
So, $n!$ is not highly composite for $n \ge 20$. Now, it remains to check if any factorials between $7!$ and $20!$ are highly composite.
EDIT: From the list of the first 1000 highly composite numbers that OP mentioned, it appears that the $149$-th largest highly composite number is $\approx 1.49 \times 10^{17}$, while $19! \approx 1.22 \times 10^{17}$. None of  the numbers $8!, 9!, \ldots, 19!$ appear in that list, so $7!$ is indeed the largest highly composite factorial.
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