Solution:

The answer is (A). It can be found by exploiting the equality $5^{n} = \frac{10^{n}}{2^{n}}$. Let $c$ be the number of decimal digits of $2^{n}$; the long division algorithm implies that the first digit of the quotient $\frac{10^{c}}{2^{n}}$ is the result of the integer division of $10^{c}$ by $2^{n}$. This result is 1, since $2^{n} \geq 7 \cdot 10^{c-1}$ and hence $2 \cdot 2^{n} \geq 14 \cdot 10^{c-1} > 10^{c}$.

Alternatively, it is possible to formally express the solution without referring to the long division algorithm. Let $x$ be the leading digit of $5^{n}$; there exist integers $k, h$ such that $7 \cdot 10^{k} \leq 2^{n} < 8 \cdot 10^{k}$ and $x \cdot 10^{h} \leq 5^{n} < (x+1) \cdot 10^{h}$. Multiplying these inequalities term by term and using the fact that $2^{n} \cdot 5^{n} = 10^{n}$ we obtain
$$7x \cdot 10^{k+h} \leq 10^{n} < 8(x+1) \cdot 10^{k+h}$$
from which in particular $10^{k+h} < 10^{n} < 10^{k+h+2}$. Clearly this is possible only if $n = k+h+1$. From the inequality $7x \cdot 10^{k+h} < 10^{n} = 10^{k+h+1}$ we then obtain $7x < 10$, which (since $x$ is a positive integer) is possible only for $x = 1$.