For which positive integers $n$ is it true that $2^{n}$ is a divisor of $\left(5^{n}-1\right)$?
We distinguish two cases: a) $n$ is odd, b) $n$ is even.

a) Write $n$ in the form $n=2k+1$! Then

$$
5^{n}-1=5^{2k+1}-1=5 \cdot 25^{k}-1=5\left(25^{k}-1\right)+4
$$

Based on the identity $a^{k}-b^{k}=(a-b)\left(a^{k-1}+a^{k-2} b+\ldots+b^{k-1}\right)$, $25^{k}-1$ is divisible by 24, and thus by 8, so $5^{n}-1$ is divisible by 4 but not by 8. Therefore, for $n \geq 3$, it is not possible that $2^{n} \mid\left(5^{n}-1\right)$. The only remaining possibility is $n=1$, which is indeed a solution.

b) Write $n$ in the form $n=2^{k} m$, where $m$ is an odd integer. Using the identity $a^{2}-b^{2}=(a-b)(a+b)$ repeatedly, we get

$$
\begin{gathered}
5^{2^{k} m}-1=\left(5^{2^{k-1} m}+1\right)\left(5^{2^{k-1} m}-1\right)=\left(5^{2^{k-1} m}+1\right)\left(5^{2^{k-2} m}+1\right)\left(5^{2^{k-2} m}-1\right)=\ldots= \\
=\left(5^{2^{k-1} m}+1\right)\left(5^{2^{k-2} m}+1\right) \cdot \ldots \cdot\left(5^{4 m}+1\right)\left(5^{2 m}+1\right)\left(5^{m}+1\right)\left(5^{m}-1\right)
\end{gathered}
$$

For any positive integer $i$, $5^{i}+1$ is even but not divisible by 4, since for $i=1$, $5^{i}+1=6$, and for $i=2$, $\left(5^{i}+1\right)=26$. Therefore, the first $k$ factors of our product are even but not divisible by 4. From the examination of case a), we know that $5^{m}-1$ is divisible by 4 but not by 8, so $2^{k+2} \mid\left(5^{n}-1\right)$, but $2^{k+3} \nmid\left(5^{n}-1\right)$. Thus, $2^{n} \mid\left(5^{n}-1\right)$ holds if and only if

$$
2^{k} m \leq k+2
$$

For $k \geq 3$, $2^{k}>k+2$, so only the cases $k=1$ and $k=2$ can apply. If $k=1$, then from (1) $2m \leq 3$, so $m=1$, and in this case $n=2$. If $k=2$, then from (1) $4m \leq 4$, so $m=1$, and $n=4$. It is easy to verify that these are indeed solutions.

In summary: $2^{n} \mid\left(5^{n}-1\right)$ only holds for $n=1,2,4$.