Show that the product of some terms of the sequence is congruent to $2$ modulo $p$. 
QUESTION: Let $p\geq 3$ be a prime number and $\mathfrak{S}=\{a_1, a_2, \cdots , a_{p-2}\}$ be a sequence of positive integers such that
  $p$ doesn't divide neither $a_k$ or $a_k^k-1$ $\forall k\in \{1,2,\cdots, k\}$. Prove that the product of some terms of the
  sequence is congruent to $2$ modulo $p$.

How to solve the problem? 
I can understand that numbers of the form $pk, pk+2; k\in\mathbb{N}$ can be excluded beforehand. Then we are left with the following $p-2$ classes of numbers, i.e, the ones congruent to : $$\bigg\{1,3,4,5,6,\cdots, \frac{p-1}{2}, -\frac{p-1}{2}, \cdots, -3,-2,-1\bigg\}$$. 
How to proceed further?
A few of the initial comments might not match. The question was initially wrongly worded. And it has been fixed. Parts of the question and my example (s) have been removed.
 A: This is a Mathematical Olympiad problem. A solution can be found here, and I give a modified version below:

Lemma: There exist $p-1$ positive integers $b_1, b_2, \ldots b_{p - 1}$ such that
a) $b_1 = 1$;
b) If $1 \leq i < j \leq p-1$, $b_i \not \equiv b_j \mod p$;
c) For $1 < i \leq j$, $b_i$ is the product of some of the $a_k$ (with $k < i$).

Proof:
Set $b_1 = 1, b_2 = a_1$.
For $2 \leq k \leq p - 2$, suppose we have defined $b_1, \ldots, b_k$ satisfying conditions a, b and c.
Consider the numbers $a_kb_1, a_kb_2, \ldots, a_kb_k$. Since $a_k \not \equiv 0 \mod p$ and all the $b_i's$ are different modulo $p$ by condition c, we know that $a_kb_i \not \equiv a_kb_j$ for $1 \leq i < j \leq k$.
Also, because $(a_k b_1) (a_k b_2) \ldots (a_k b_k) = a_k^k (b_1 \ldots b_k) \not \equiv b_1 \ldots b_k \mod p $ (because $a_k^k \not \equiv 1 \mod p$), we also know that $(a_kb_1, a_kb_2, \ldots, a_kb_k)$ is not a permutation of $(b_1, \ldots, b_k)$ modulo $p$, so that there must exist some $j$ such that $a_kb_j \not \equiv b_i \mod p$ whenever $1 \leq i \leq k$.
Set $b_{k+1} = a_k b_j$, and it's trivial to check that conditions a,b and c are still satisfied.

To finish the proof, let $b_1, \ldots, b_{p-1}$ be as in the Lemma. Because $a_k \not \equiv 0 \mod p$ for each $k$, we know that $b_k \not \equiv 0 \mod p$, so that $(b_k \mod p) \in \{1, \ldots, p-1\}$. By the pidgeonhole principle, there must exist some $k$ such that $b_k \equiv 2 \mod p$; since $b_1 = 1$, we have $k > 1$ and, by condition c in the Lemma, $b_k$ is the product of some of the $a_j$. We are done.

Obs: 
1) The choice of $2$ in the problem is arbitrary. There exists some product that is congruent to $m$ modulo $p$ for any $m \in \{2, 3, \ldots, p -1\}.$
2) If we look at the proof of the Lemma, we might get the impression that if the sequence of $a_k's$ were longer (say infinite), we could go on forever and get a sequence of $b_k$ that is as long as we'd like. But this obviously cannot be true by condition b. Is this a contradiction?
No, because by Fermat's Little theorem, the condition $a_k^k \not \equiv 1 \mod p$ cannot be true for $k = p -1$ if $a_k \not \equiv 0 \mod p$. Therefore, we could not have a longer sequence of $a_k$ satisfying both conditions.
