When calculators probably can't help

$\dfrac{n!}{k} = (2\times3\times5\times7)^{2+3+5+7} = (2\times3\times5 \times 7) ^{17}$

The smallest $n!$ must be the smallest number which has $7^{17}$ as a factor.

The power of $7$ in $n!$ can be found with: $p_7 = \displaystyle \sum_{k=1}^\infty {\left \lfloor \frac{n}{7^k} \right \rfloor}$ , where $\lfloor \space \rfloor$ is the greatest integer function.

For $p_7 = 17$ , $n=\boxed{105}$ .

The expression suggests that $n!$ should have at least $17$ factors of $2$ , $3$ , $5$ , and $7$ . Since $7$ occurs relatively rarer than the other digits, then it has to be assured that $n!$ contains at least $17$ factors of $7$ .

Therefore, we have to find $n$ such that the total factors of $7$ from $1$ to $n$ amount to $17$ . We can say that $n$ may be equal to $17 \times 7 = 119$ if we consider every multiple of $7$ occurring from $1$ to $n$ inclusive. However, we failed to consider the fact that $49$ and $98$ each contribute to two factors of $7$ . Thus, we count $49$ and $98$ twice and the multiples included will only be $15$ . Thus the answer, $15 \times 7 = \boxed {105}$