**EDIT:** here's a better approach; in particular, the formula given in my original answer for $d(n!)$ is completely incorrect as pointed out by Gerhard Paseman.

Divide the primes in $[1,n]$ into the "small primes" $p \le \sqrt{n}$ and the "large primes" $p \ge \sqrt{n}$. Denote by $\pi[a,b]$ the number of primes in the interval $[a,b]$.

For any two small primes $p \neq q$ we have $v_p(n!) \neq v_q(n!)$ - just compare the terms $\lfloor n/p \rfloor$ and $\lfloor n/q \rfloor$. For any large prime $p$ we have $v_p(n!) = \lfloor n/p \rfloor$. In addition, we have $v_p(n!) < n$ for any (small or large) prime $p$.

Now the number of divisors of $n!$ is $\prod_{p \le n\text{ prime}} 1+v_p(n!)$. We split this product into small and large primes.

The product of the small-prime factors is the product of $\pi[1,\sqrt{n}]$ distinct integers between $\sqrt{n}$ and $n$, and hence divides $n!$. In fact, aside from $1+v_2(n!)$, the small-prime factors all lie between $\sqrt{n}$ and $n/2$, and so their product divides $(n/2)!$.

For the large primes a bit more work is needed. We count the number of primes with $\lfloor n/p \rfloor = k$: this is just (up to boundary terms) $\pi\left[\frac{n}{k+1}, \frac{n}{k}\right].$ So the product of the factors for small primes looks something like $$P = 2^{\pi\left[\frac{n}{2},n\right]} \cdot 3^{\pi\left[\frac{n}{3},\frac{n}{2}\right]} \cdot 4^{\pi\left[\frac{n}{4},\frac{n}{3}\right]} \cdots \sqrt{n}^{\pi\left[\sqrt{n},\sqrt{n}+1\right]}.$$

We claim that for large $n$, this product divides $(n/2)!$. We have $$v_2(P) = \pi[n/2,n] + 2\pi[n/4,n/3] + \pi[n/6,n/5] + 3\pi[n/8,n/7] + \cdots.$$ We know that the partial sums of the sequence $\{v_2(2k)\}_{k \ge 1} = \{1, 2, 1, 3, 1, 2, 1, \cdots\}$ are always less than $2k$, so provided $\log n$ is large enough that PNT asymptotics are reasonably valid (e.g., $\pi[n/3, n/2] \ge \pi[n/4, n/3]$) we have $$v_2(P) \le \pi[1,n] \approx \frac{n}{\log n} < v_2((n/3)!) \approx \frac{n}{3}$$ for large enough $n$. An analogous argument gives $v_p(P) \le \pi[1,n/(p-1)]$ which is again eventually less than $v_p((n/3)!)$.

To finish, we note that our total product is at most $(n/2)!(n/3)!\cdot (1+v_2(n!))$. Either $1 + v_2(n!)$ is a prime greater than $n/2$ or it is divisible by $(n/6)!$ for large enough $n$. Finally, we have $(n/2)!(n/3)!(n/6)!$ divides $n!$ by an elementary modular arithmetic argument.

This doesn't give any explicit bound on $n$, but it shows why such a result must be true for large enough $n$. With a fair bit more analytic number theory and cleverness, an explicit bound could presumably be found.

~~ Vinoth is right: this is suitable for talented high school students. Here's a proof sketch.~~

~~
~~$d(n)$, the number of divisors, is multiplicative. I'll write $\pi[a,b]$ to denote the number of primes $p$ with $\lfloor a+1 \rfloor \le p \le \lfloor b \rfloor$.

The number of divisors of $n!$ is then $$d(n!) = 2^{\pi[\sqrt{n},n]} \cdot 3^{\pi[\sqrt[3]{n},\sqrt{n}]} \cdot 4^{\pi[\sqrt[4]{n},\sqrt[3]{n}]} \cdots$$

The power of a prime $p$ dividing $n!$ is just $$v_p(n!) = \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \cdots.$$

So one only needs to establish inequalities of the form

$$\pi[\sqrt{n},n] + 2\pi[\sqrt[4]{n},\sqrt[3]{n}] + \cdots \le \left\lfloor \frac{n}{2} \right\rfloor + \left\lfloor \frac{n}{4} \right\rfloor + \left\lfloor \frac{n}{8} \right\rfloor + \cdots,$$

$$\pi[\sqrt[3]{n},\sqrt{n}] + \pi[\sqrt[6]{n},\sqrt[5]{n}] + 2\pi[\sqrt[9]{n},\sqrt[8]{n}] + \cdots \le \left\lfloor \frac{n}{3} \right\rfloor + \left\lfloor \frac{n}{9} \right\rfloor + \left\lfloor \frac{n}{27} \right\rfloor + \cdots,$$

and so on for primes $p \ge 5$.

If we could establish the following general inequality for $m \le n$, we'd be done:

$$\sum_{k \ge 1} \pi\left[\sqrt[km]{n},\sqrt[km-1]{n}\right] \le \left\lfloor \frac{n}{m} \right\rfloor.$$

~~The left-hand side is bounded above by $\pi[1,\sqrt[m-1]{n}]$, and the inequality $$\pi[1,\sqrt[m-1]{n}] \le \left\lfloor \frac{n}{m} \right\rfloor$$ is straightforward ($\pi[a,b] \le b - a + 1$) for large $n$ unless $m = 2$. In this case we can use an explicit PNT estimate of the form $\pi[1,n] \le n/C\log n$ for some $1 < C < 2$, or just use an estimate of the form $\pi(n) \le 2 + \frac{n}{3}$, which is true for all $n$ by arithmetic modulo 6. ~~