show this $(x_{1}+x_{2}+\cdots+x_{n})^p\equiv x^p_{1}+x^p_{2}+\cdots+x^p_{n}\pmod p$ Let $x_{1},x_{2},\cdots,x_{n}$ are postive integers,and $p$ is prime number,such
$$(x_{1}+x_{2}+\cdots+x_{n})^p\equiv x^p_{1}+x^p_{2}+\cdots+x^p_{n}\pmod p$$
Can this problem be solved using pre-university mathematics?
I am talking about elementary number theoretical solutions. Do they exist?
 A: If you know how to prove
$$(a_1+a_2)^p \equiv a_1^p + a_2^p\pmod{p} $$
(through Fermat's little theorem) you just have to apply induction since, for instance:
$$ (a_1+a_2+a_3)^p = (a_1+(a_2+a_3))^p \equiv a_1^p+(a_2+a_3)^p \equiv a_1^p+a_2^p+a_3^p\pmod{p}.$$
A: $$(****)(x_{1}+x_{2}+\cdots+x_{n})^p\equiv \\x^p_{1}+x^p_{2}+\cdots+x^p_{n}+\sum_{\large{k_i=1,2,3,...,p-1\\k_1+k_2+...+k_n=p}}^{}\dfrac{p!}{k_1!k_2!...k_n!}x_1^{k_1}x_2^{k_2}...x_n^{k_n}\pmod p$$ now :note that $p$ is prime so $$\dfrac{p!}{k_1!k_2!...k_n!}=\dfrac{p(p-1)!}{k_1!k_2!...k_n!}\\k_1,k_2,...k_n<p \to \\k_1!k_2!...k_n!|(p-1)! \\\ \to \\
\dfrac{p(p-1)!}{k_1!k_2!...k_n!}=p.\dfrac{(p-1)!}{k_1!k_2!...k_n!}=pQ$$ npw look again to (****)$$(x_{1}+x_{2}+\cdots+x_{n})^p\equiv \\\equiv x^p_{1}+x^p_{2}+\cdots+x^p_{n}+\sum_{\large{k_i=1,2,3,...,p-1\\k_1+k_2+...+k_n=p}}^{}pQ.x_1^{k_1}x_2^{k_2}...x_n^{k_n}\pmod p\\\equiv x^p_{1}+x^p_{2}+\cdots+x^p_{n}+p\times\sum_{\large{k_i=1,2,3,...,p-1\\k_1+k_2+...+k_n=p}}^{}Q.x_1^{k_1}x_2^{k_2}...x_n^{k_n}\pmod p\\\equiv
x^p_{1}+x^p_{2}+\cdots+x^p_{n}+0\times\sum_{\large{k_i=1,2,3,...,p-1\\k_1+k_2+...+k_n=p}}^{}Q.x_1^{k_1}x_2^{k_2}...x_n^{k_n}\pmod p\\x^p_{1}+x^p_{2}+\cdots+x^p_{n}\pmod p$$
