7. Given odd prime numbers $x, y, z$ satisfying
$$
x \mid \left(y^{5}+1\right), y \mid \left(z^{5}+1\right), z \mid \left(x^{5}+1\right) \text {. }
$$

Find the minimum value of the product $x y z$. (Cheng Chuanping, problem contributor)
7. Let's assume $x$ is the minimum of $x, y, z$.
(1) If $x=3$, then
$$
3^{5}+1=244=2^{2} \times 61 \Rightarrow z=61.
$$

Since $3 \mid (y^{5}+1)$, we have $y \equiv -1 \pmod{3}$.
Clearly, $5 \nmid (61^{5}+1)$.
After calculation, we find $11 \mid (61^{5}+1)$.
Thus, $y_{\text{min}}=11$.
Therefore, $(x y z)_{\min}=3 \times 11 \times 61=2013$.
This indicates that the smallest set of three consecutive odd primes greater than 2013 is $11 \times 13 \times 17=2431$.
Hence, we only need to consider the cases where $x=5$ or $x=7$.
(2) If $x=5$, note that,
$$
5^{5}+1=3126=2 \times 3 \times 521.
$$

Since $5 \times 521 > 2013$, this does not meet the requirement.
(3) If $x=7$, we estimate the value of $7^{5}+1$.

Since $7 \mid (y^{5}+1)$, we have $y \equiv -1 \pmod{7}$.
If $y=13$, then $13 \mid (z^{5}+1)$.
Thus, $z \equiv -1 \pmod{13}$.
Hence, $z_{\text{min}}=103$, and $(x y z)_{\min} > 2013$.
In summary, the minimum value of $x y z$ is 2013.