Prove that there exist primes $p_{i}$ such that $\prod_{i=1}^{k}p_{i} \mid \sum_{i=1}^{k}(p_{i})^{a_{i}}$ If $k\geq 3$ is a given positive integer, prove that there exist  prime numbers $p_{1}<p_{2}<\cdots<p_{k}$ and positive integers $a_{1},a_{2},\cdots,a_{k}$, such that
$$p_{1}p_{2}\cdots p_{k} \mid (p_{1})^{a_{1}}+(p_{2})^{a_{2}}+\cdots+(p_{k})^{a_{k}}$$
I need to prove
$$p_{i} \mid (p_{1})^{a_{1}}+(p_{2})^{a_{2}}+\cdots+(p_{i-1})^{a_{i-1}}+(p_{i+1})^{a_{i+1}}+\cdots+(p_{k})^{a_{k}}\quad\forall i=1,2,\ldots,k$$
from Fermat's little theorem, we need prove there exist $a_{1},a_{2},\ldots,a_{n}$ s.t.
\begin{align} 
a_{1} &= 0 \pmod {(p_{i}-1)} &(i &\neq 1) \\
a_{2} &= 0 \pmod {(p_{i}-1)} &(i &\neq 2) \\
&\;\;\vdots \\
a_{k} & = 0 \pmod {(p_{i}-1)} & (i &\neq k)
\end{align}
it seem use CRT, but in fact we can't use this theorem, because this problem not  such Chinese remainder theorem condition, so How to prove this problem
 A: Choose $q_{1}$ to be a prime that divides $k-1$
By lemma 3 (proved below) exists $j_{1}$ so that $q_{1}^{j_{1}}+k-2$ is a multiple of some prime $q_{2}>q_{1}$. $\text{ }$ By lemma 3 there exists $j_{2} $ so that $q_{1}^{j_{1}}+q_{2}^{j_{2}(q_{1}-1)}+k-3$ is a multiple of some prime $q_{3}>q_{2}$...(doing this process many times)...by lemma 3 there exists $j_{k-1}$ so that $$q_{1}^{j_{1}}+q_{2}^{j_{2}(q_{1}-1)}+...+q_{k-1}^{j_{k-1}(q_{1}-1)(q_{2}-1)...(q_{k-2}-1)}$$ is a multiple of some prime $q_{k}>q_{k-1}$. Thus $$q_{1}^{j_{1}}+q_{2}^{j_{2}(q_{1}-1)}+...+q_{k-1}^{j_{k-1}(q_{1}-1)(q_{2}-1)...(q_{k-2}-1)}+q_{k}^{(q_{1}-1)(q_{2}-1)...(q_{k-2}-1)(q_{k-1}-1)} \equiv 0 \mod q_{1}q_{2}...q_{k}$$
We will prove Lemma 3, an important ingredient for our proof below.

$\textbf{Definition:}$ Given a prime $p$ and a positive integer $n$. $M(p,n)$ is the unique non-negative integer such that $p^{M(p,n)}|n$ but $p^{M(p,n)+1} \not | n$
$\textbf{Lemma 1:}$ If
$$a \in \mathbb{N}$$
$$b \in \mathbb{N}; b > 1$$
$$c \in \mathbb{Z}; c \neq 0$$
then given a prime $p$, there exists a non-negative integer $e_{1}$, natural numbers $u, v$ such that for all $t \in \mathbb{N}\cup \{0\}$ we have
$$M(p,ab^{u+vt} + c) = e_{1}$$
$\textbf{Proof:}$
If $p|c$ then there exists $h$ sufficiently large so that $M(p,ab^{h} + c) = M(p,ab^{h+1} + c)=....$ thus it sufficies to take $u = h$ and $v = 1$. If $p \not | c$ but $p| a$ or $b$ then it sufficies to take $u = 1, v = 1$. If $p \not | a, b$ or $c$ then suppose $a b^{s} + c > 1$ we can take $$u = s$$ $$v = (p-1)s_{1}^{M(p,a b^s +c)}$$ $$e_{1} = M(p,a b^s +c)$$
$\textbf{Lemma 2:}$ If
$$a \in \mathbb{N}$$
$$b \in \mathbb{N}; b > 1$$
$$c \in \mathbb{Z}; c \neq 0$$
then given a set of primes $\{s_{1},...,s_{k}\}$, there exists non-negative integers $e_{1},...,e_{k}$ and natural numbers $u_{k},v_{k}$ such that for all $1 \leq j \leq k$, $t \in \mathbb{N} \cup \{0\}$ we have
$$M(s_{j}, a b^{u_{k}+v_{k}t}+c) = e_{j}$$
$\textbf{Proof:}$
By the above lemma there exists $v_{1} \in \mathbb{N}\cup \{0\}, r_{1},s_{1} \in \mathbb{N} $ so that for all $t \in \mathbb{N}\cup \{0\}$ we have $M(s_{1}, a b^{r_{1}+s_{1}t}+c) = e_{1}$, for some fixed $e_{1} \in \mathbb{N}\cup \{0\}$. Now $ a b^{r_{1}+s_{1}t}+c = (ab^{r_{1}})(b^{s_{1}})^t+c$ thus we can continually apply Lemma 1 to prove this lemma.
$\textbf{Lemma 3:}$ If
$$a \in \mathbb{N}$$
$$b \in \mathbb{N}; b > 1$$
$$c \in \mathbb{Z}; c \neq 0$$
then given a set of primes $\{s_{1},...,s_{k}\}$. There exists a prime $q \not \in \{s_{1},...,s_{k}\}$ and an integer $s$ so that $q | ab^s+c$.
$\textbf{Proof:}$ Suppose that this is not the case. Note that by lemma two $M(s_{j},ab^{u_{k}+v_{k}t}+c)$ is fixed for a fixed $j$ and variable $t \in \mathbb{N}\cup \{0\}$. Thus we must have $ab^{u_{k}+v_{k}0}+c = ab^{u_{k}+v_{k}1}+c = ...$ which is impossible.
