Any implications?

If $8! + k$ is prime, $k$ must be greater than $8$ and cannot be divisible by any multiples of $2 \rightarrow 8$ ( this problem will illustrate why).

Thus, $k$ must have prime factors $\geq 11$ . Testing the combinations of multiplied primes $\geq 11$ , we quickly find our number:

$\begin{aligned} 8! + (11 \times 11) & = 37 \times 1093 \rightarrow \text{composite} \\ 8! + (11 \times 13) & = 43 \times 941 \rightarrow \text{composite} \\ 8! + (13 \times 13) & = 19 \times 2131 \rightarrow \text{composite} \\ 8! + (11 \times 17) & = 1 \times 40507 \rightarrow \text{prime} \end{aligned}$

Thus, the minimum composite $k$ is $11 \times 17 = \boxed{187}$ .

$8!=40320=2^7\cdot 3^2\cdot 5\cdot 7$

Suppose that $k$ is of the form $2^a\cdot 3^b\cdot 5^c\cdot 7^d\cdot n$ $(a,b,c,d,n\in\mathbb{N})$ with at least one of $a,b,c,d$ non-zero. Wlog assume $a\neq 0$ , then

$8!+k=2(20160+2^{a-1}\cdot 3^b\cdot 5^c\cdot 7^d\cdot n)$

Which is not prime. With a similar reasoning also non-zero $b,c$ or $d$ make $8!+k$ a composite number. So, according to the second constraint, $k$ does not containt factors of $2,3,5$ or $7$ in its prime factorization

We now search for the least possible composite $k$ that satisfies the second constraint. Proceeding in ascending order:

$11^2,11\cdot 13,13^2,11\cdot 17,13\cdot 17,\ldots$

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$8!+11^2=40441=37\cdot 1093$

$8!+11\cdot 13=40463=43\cdot 941$

$8!+13^2=40489=19\cdot 2131$

$8!+11\cdot 17=40507 \color{#D61F06}\text{ prime}$

Hence the least possible $k$ is

$k=11\cdot 17=\boxed{187}$