Saudi Arabian Mathematical Competitions, 2015

Assume that there exists an integer $k$ such that $0 \leq k \leq p^{2}$ and $p^{2}$ divides $k(k+1)(k+2) \cdots(k+p-3)-1$. Because $p$ does not divide any of the $p-2$ consecutive integers $k,(k+1),(k+2), \ldots,(k+p-3)$, either $k \equiv 1$ or $2 \bmod p$. If $k \equiv 2 \bmod p$ then, by Wilson's theorem,
$$k(k+1)(k+2) \cdots(k+p-3)-1 \equiv (p-1)!-1 \equiv -2 \not \equiv 0 \bmod p.$$
We deduce that $k \equiv 1 \bmod p$.

On the other hand, we know from Wilson's theorem that $(p-2)!-1 \equiv 0 \bmod p$. This means that there exists a unique integer $i_{0} \in \{1, \ldots, p\}$ such that
$$(p-2)!-1 \equiv i_{0} p \bmod p^{2}.$$
Let $k = i p + 1$ for some integer $i \in \{0,1, \ldots, p-1\}$. We have
$$\begin{aligned} k(k+1)(k+2) \cdots(k+p-3)-1 &\equiv (i p+1)(i p+2) \cdots(i p+p-2)-1 \\ &\equiv (p-2)!-1 + i p \sum_{j=1}^{p-2} \frac{(p-2)!}{j} \bmod p^{2} \end{aligned}$$
and
$$\sum_{j=1}^{p-2} \frac{(p-2)!}{j} \equiv (p-2)! \sum_{j=1}^{p-2} j^{-1} \equiv \sum_{j=1}^{p-2} j \equiv \frac{(p-2)(p-1)}{2} \equiv 1 \bmod p$$
We deduce that
$$k(k+1)(k+2) \cdots(k+p-3)-1 \equiv (i_{0}+i) p \bmod p^{2}.$$
This proves that $k(k+1)(k+2) \cdots(k+p-3)-1 \equiv 0 \bmod p^{2}$ only for $k = (p-i_{0}) p + 1$.