16. 2. $47 \star \star$ For some natural numbers $n$, the first digit of the numbers $2^{n}$ and $5^{n}$ is the same. What are these first digits?

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Since $2^{5}=32,5^{5}=3125$, we know that the leading digit can be 3. Clearly, we only need to consider $n>3$. Suppose the leading digit of both $2^{n}$ and $5^{n}$ is $a$, and they are $s+1$ and $t+1$ digits long, respectively, then
$$
a \cdot 10^{s} < 2^{n} < (a+1) \cdot 10^{s}, \quad a \cdot 10^{t} < 5^{n} < (a+1) \cdot 10^{t}.
$$
Multiplying these two inequalities, we get
$$
a^{2} \cdot 10^{s+t} < 2^{n} \cdot 5^{n} = 10^{n} < (a+1)^{2} \cdot 10^{s+t}.
$$
This implies
$$
a^{2} < 10^{n-s-t} < (a+1)^{2}.
$$
Since $10^{n-s-t}$ is a power of 10, it must be an integer. Given that $a$ is an integer, it must be that $a=3$.