7. Let $x_{i}(i=1,2,3,4)$ be real numbers, and $\sum_{i=1}^{4} x_{i}=\sum_{i=1}^{4} x_{i}^{7}=0$, find the value of the following expression $u=x_{4}\left(x_{4}+x_{1}\right)\left(x_{4}+\right.$ $\left.x_{2}\right)\left(x_{4}+x_{3}\right)$.
7. Regarding $x_{i}$ as indeterminate elements, then $\sigma_{1}=S_{1}=S_{7}=0$. By Newton's formula for $k \geqslant n$, $S_{2}-\sigma_{1} S_{1}+2 \sigma_{2}=0$. Thus, $S_{2}=-2 \sigma_{2}$. By $S_{3}-\sigma_{1} S_{2}+\sigma_{2} S_{1}-3 \sigma_{3}=0$, we have $S_{3}=3 \sigma_{3}$.
Similarly, $S_{4}=2 \sigma_{2}^{2}-4 \sigma_{4}, S_{5}=-5 \sigma_{2} \sigma_{3}, S_{7}=7\left(\sigma_{2}^{2}-\sigma_{4}\right) \sigma_{3}$. From $S_{7}=0$, we get $\sigma_{3}=0$ or $\sigma_{4}=\sigma_{2}^{2}$.
If $\sigma_{3}=0$, then $f(t)=\left(t-x_{1}\right)\left(t-x_{2}\right)\left(t-x_{3}\right)\left(t-x_{4}\right)=t^{4}+\sigma_{2} t^{2}+\sigma_{4}$.

From this, we know that the 4 roots of $f(x)$ are two pairs of opposite numbers, so $x_{4}$ is the opposite number of one of $x_{1}, x_{2}, x_{3}$, thus $u=0$.
If $\sigma_{4}=\sigma_{2}^{2}$, then $S_{4}=-2 \sigma_{2}^{2} \leqslant 0$, leading to $S_{4}=0$, thus $x_{1}=x_{2}=x_{3}=x_{4}=0$. Hence, $u=0$.