$\displaystyle \lim_{n \to \infty} \dfrac{10^n}{n!}$
$= \dfrac{\overbrace{10 \times 10 \times 10 \times \ldots \times 10 \times 10}^{n \text{ times}}}{1 \times 2 \times 3 \times \dots \times (n-1) \times n}$
$ =\overbrace{\frac{10}{1} \cdot \frac{10}{2} \dots \frac{10}{10}}^{\approx 2755} \cdot \underbrace{\frac{10}{11} \dots \frac{10}{100} \cdot \frac{10}{101}}_{\ll 1} \dots \overbrace{\frac{10}{10000} \cdot \frac{10}{10001}}^{\lll 1} \dots \small \text{ goes forever}$
Now we can see that after the $\frac{10}{10}$ term, all subsequent terms are $<1$, and keep decreasing. As we increase the value of $n$ the product will get close to $0$.
So as $n \to \infty$ $\dfrac{10^n}{n!}\to 0$.
Hence, the answer is option A.