Let $p>n+1$ then $p^2\not | x^3+1$, where $1 \leq x \leq n$, because $p^2>x^2-x+1$ and $p \not | x+1$ So if $p^3|(1^3+1)(2^3+1)...((n-1)^3+1)(n^3+1)$ then exists $1<a<b<c \leq n$ that $p|a^2-a+1,p|b^2-b+1,p|c^2-c+1$ But then $p|b^2-b-a^2+a=(b-a)(b+a-1) \to p|b+a-1$ similar $p|c+a-1 \to p|c-a \to $ contradiction

Not much different from the above post. $[i]$ If $p^3\mid k^3+1$ for some $k\leq n$; then since $(k+1,k^2-k+1)=1$ (unless $p=3$, which can be manually handled), we have either $p^3\mid k+1$ or $p^3\mid k^2-k+1$, both of which gives us the desired result. $[ii]$ Next, if there exists a $k$ such that $p^2\mid k^3+1$, then we still have the same contradiction (as if $p^2\mid k+1 \implies p^2\leq k+1\leq n+1$; and if $p^2\mid k^2-k+1$ then $p^2 \leq k^2-k+1<k^2\leq n^2 \implies p\leq n$. $[iii]$ Finally, there must exist three numbers $1\leq k_1<k_2<k_3\leq n$ such that $p\mid k_i^2+1$, for each one of them. Clearly, if $p\mid k_i+1$ holds for any one of them, we have the result; so assume $p\mid k_i^2-k_i+1$ for each $i=1,2,3$. If $p>n+1$; then the polynomial $f(x)=x^2-x+1$ has at least $3$ incongruent roots in $\pmod{p}$, contradicting with Lagrange's theorem.

idon't care about the bad writeup but I beat adi

@Above adi beat you by a second blud dont be cappin anyway solution: Assume FTSOC $p > n + 1$. If $p \mid (1 + 1)(2 + 1)\dots(n + 1)$, clearly $p \leq n + 1$, so assume not. Then $p^3 \mid (1^2 - 1 + 1)\dots(n^2 - n + 1)$. Note that $p^2 > k^2 - k + 1$ for any particular $1 \leq k \leq n$ since $p^2 > k^2$, so at least three pairwise distinct $x, y, z$ must exist such that $p \mid x^2 - x +1, p \mid y^2 - y + 1, p \mid z^2 - z + 1 \implies (2x - 1)^2 \equiv (2y - 1)^2 \equiv (2z - 1)^2 \mod{p}$. $2x - 1 \equiv 2y - 1 \mod{p} \implies x \equiv y\mod{p}$, contradiction as $x, y < p$, so $p \mid 2x - 1 + 2y - 1 \implies p \mid x + y - 1$, but $x + y - 1 < 2p \implies x + y = p + 1$. So, $x + y = p + 1, y + z = p + 1, x + z = p + 1 \implies x = y = z$, contradiction, hence $p > n + 1$