Solution:

The answer is 12. Let $N=\frac{n !}{(n-6) !}=(n-5)(n-4)(n-3)(n-2)(n-1) n$.

Let us first look for the maximum $q$ such that $q$ divides $N, \forall n$. Among six consecutive integers, three are divisible by 2, and among these three, one is certainly divisible by 4. Hence $2^{4}$ divides $N$. Among six consecutive integers there are two integers divisible by 3. Hence $3^{2}$ divides $N$. Among six consecutive integers, only one of them is divisible by 5. Hence 5 divides $N$. Therefore
$$q \geq 2^{4} \cdot 3^{2} \cdot 5$$
Since we are looking for the maximum $q$ such that $q$ divides $N, \forall n$, let us consider $n_{1}=7, n_{2}=13$. Certainly $q$ divides both $N_{1}$ and $N_{2}$, from which it follows that
$$q \text{ divides } \operatorname{MCD}\left(N_{1}, N_{2}\right)$$
One finds that $N_{1}=2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7, N_{2}=8 \cdot 9 \cdot 10 \cdot 11 \cdot 12 \cdot 13$. Then $\operatorname{MCD}\left(N_{1}, N_{2}\right)=2^{4} \cdot 3^{2} \cdot 5$. Now, from 1 and 2 we deduce that $q=2^{4} \cdot 3^{2} \cdot 5$. Wishing to extract from $q$ the largest square, we find that $k^{2}=2^{4} \cdot 3^{2}$, that is $k=12$.