Here's an approach by thinking about orders of magnitude. Factorials can be approximated well using Stirling's approximation but even the crude bounds in this answer deal with the problem, and a tougher generalisation, quite quickly.
We seek $n! = 7! \times 5! \times 3!$ and, given how quickly factorials grow, it's reasonable to think $n$ cannot be much more than $7$. In fact there are no solutions known for $a! = b! \times c! \times \cdots$ with $a > b + 3$ (where without loss of generality $b \ge c \ge \cdots$).
With $n = 7 + k$, and clearly $k \ge 1$, consider solving $(7+k)! = 7! \times 5! \times 3!$. We need
$$ (7+1)(7+2)\cdots (7+k) = 5! \times 3! $$
so an obvious inequality is
$$ (7+1)^k \le 5! \times 3! \le (7+k)^k $$
and if $k>1$ these bounds become strict. Indeed it's clear that $k \ne 1$ since $(7+1)^1 = 8 \ll 5! \times 3!$. We could also make a better estimate using the arithmetic mean of $7+1, 7+2, \dots, 7+k$ which is $7 + (k+1)/2$, so that $(7 + (k+1)/2)^k \approx 5! \times 3!$.
The AM-GM inequality then gives that $(7 + (k+1)/2)^k > (7+1)\cdots (7+k)$ for $k>1$. So we can use the tighter inequality
$$ 7+1 \lt \sqrt[k]{5! \times 3!} \lt 7+ \frac{k+1}{2} $$
and since $5! \times 3! = 120 \times 6 = 720$ is quite small, we will only need to consider the first few roots. If a $k$-th root lies within plausible bounds then we will check if $(7+1)\cdots (7+k)=720$.
For $k=2$, $\sqrt{5! \times 3!} = 26.83\dots > 7 + 3/2$ is no good.
For $k=3$, $\sqrt[3]{5! \times 3!} = 8.96\dots \in (7 + 1, 7 + 4/2)$ is plausible. In other words, we can see $5! \times 3! \approx 9^3$ but need to check if it exactly equals $8 \times 9 \times 10$. Both are $720$ so we have found a solution $10! = 7! \times 5! \times 3!$.
We could stop there as if this question has a solution it is clearly unique, but the next step is illustrative had the question been slightly different and we had not found a solution at $k=3$. For $k=4$, $\sqrt[4]{5! \times 3!} = 5.18\dots < 7 + 1$ is no good, and it's clear higher roots will take us even further below the lower bound. So if we hadn't found a solution already for lower $k$, there wouldn't be one at all.
This method is "overkill" in some ways, but the calculations aren't terribly difficult — tables of logarithms have been around for over four centuries. What's nice is that it easily generalises to the more general question of finding integer $a$ and $b$ such that
$$a! = b! \times 5! \times 3!$$
for which we obtain the inequality
$$ b+1 \le \sqrt[k]{5! \times 3!} \le b + \frac{k+1}{2} $$
with $a = b + k$, $k \ge 1$. Again the inequality becomes strict if $k > 1$.
For $k = 1$ we have $b + 1 = 720$ so $b = 719$ and $a = b + k = 720$. So we find the solution $720! = 719! \times 5! \times 3!$.
For $k=2$, $\sqrt{5! \times 3!} = 26.83\dots \notin (b + 1, b + 3/2)$ as that would require a fractional part between $0$ and $0.5$.
For $k=3$, $\sqrt[3]{5! \times 3!} = 8.96\dots \in (b + 1, b + 4/2)$ is plausible. It would require $6.96\dots < b < 7.96\dots$ and hence $b = 7$. We have already confirmed $8 \times 9 \times 10 = 720$, so have recovered the solution $10! = 7! \times 5! \times 3!$.
For $k=4$, $\sqrt[4]{5! \times 3!} = 5.18\dots \in (b + 1, b + 5/2)$ would require $2.68\dots < b < 4.18\dots$ so $b=3$ or $b=4$. But neither work, since $4 \times 5 \times 6 \times 7 > 720$ and so clearly $5 \times 6 \times 7 \times 8 > 720$ also.
For $k=5$, $\sqrt[5]{5! \times 3!} = 3.72\dots \in (b + 1, b + 6/2)$ would require $0.72\dots < b < 2.72\dots$ so $b=1$ or $b=2$. For $b = 1$, we need to check $2 \times 3 \times 4 \times 5 \times 6 = 720$, which it does. So we have found the solution $6! = 1! \times 5! \times 3!$. For $b = 2$, clearly the product $3 \times 4 \times 5 \times 6 \times 7 > 720$.
For $k=6$, $\sqrt[6]{5! \times 3!} = 2.99\dots \in (b + 1, b + 7/2)$ would require $-0.51 \dots < b < 1.99 \dots$ so $b=0$ or $b=1$. For $b=0$, we need to check $1 \times 2 \times 3 \times 4 \times 5 \times 6 = 720$, which it does. So we have found the solution $6! = 0! \times 5! \times 3!$. For $b = 1$, clearly the product $2 \times 3 \times 4 \times 5 \times 6 \times 7 > 720$.
For $k>6$, we know $\sqrt[k]{5! \times 3!} < 3$ and so it will only lie in $(b + 1, b + (k+1)/2)$ for non-negative $b$ if $b=0$ or $b=1$. Either way the products $(b+1)(b+2) \cdots (b+k)$ will be too large, as they will be bigger than in the case for $k=6$.
Overall we obtain the exact solutions $720! = 719! \times 5! \times 3!$, $10! = 7! \times 5! \times 3!$, $6! = 1! \times 5! \times 3!$ and $6! = 0! \times 5! \times 3!$. Examining our results for $k=2$ and $k=4$ more closely, our method has also found the "near(ish) misses":
$$\frac{27!}{25!5!3!} = 0.975; \frac{28!}{26!5!3!} = 1.05; \frac{6!}{5!3!2!} = 0.5; \frac{7!}{5!3!3!} = 1.1\dot{6} $$