Let $C_n$ denote the number of (monic) degree $n$ polynomials $P(x) = \prod_{t=1}^{n}(x-t) \pmod {n!}$ such that $n! | P(x)$ for all integers $x$. Here are the first four examples:
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Polynomials of degree 2 (mod 2) that are divisible by 2 for all x
x^2+x
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Polynomials of degree 3 (mod 6) that are divisible by 6 for all x
x^3 + 5*x
x^3 + 3*x^2 + 2*x
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Polynomials of degree 4 (mod 24) that are divisible by 24 for all x
x^4 + 2*x^3 + 11*x^2 + 10*x
x^4 + 2*x^3 + 23*x^2 + 22*x
x^4 + 6*x^3 + 11*x^2 + 6*x
x^4 + 6*x^3 + 23*x^2 + 18*x
x^4 + 10*x^3 + 11*x^2 + 2*x
x^4 + 10*x^3 + 23*x^2 + 14*x
x^4 + 14*x^3 + 11*x^2 + 22*x
x^4 + 14*x^3 + 23*x^2 + 10*x
x^4 + 18*x^3 + 11*x^2 + 18*x
x^4 + 18*x^3 + 23*x^2 + 6*x
x^4 + 22*x^3 + 11*x^2 + 14*x
x^4 + 22*x^3 + 23*x^2 + 2*x
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Trivially, we have
$C_1 = 1$ (since $x$ is the only such polynomial)
From the tables we have,
$C_2 = 1, C_3 = 2, C_4 = 12$.
For $C_5$, manual computations showed there are $288$ such polynomials.
For general $n$, a counting technique can be used where all possible polynomials $P(x)$ $\mod {}$ prime powers $p^k | n!$ are considered and using the Chinese Remainder Theorem.
The answer seems to be that $C_n = \prod_{t=1}^{n-1} t!$, but how would this be proven?