Prove (1) When $n=2 p_{k}$,
Since $k>p_{k}$, then $p_{k}>2 p_{k}-k=n-k$, so, $p_{k} \nmid(n-k)!$, hence $n \nmid(n-k)!$.
(2) When $n>2 p_{k}$,

Since $n$ is a composite number, let $n=a b(2 \leqslant a \leqslant b)$, if $a \geqslant 3$.
( i ) $a \neq b$, then $n>2 p_{k} \geqslant \frac{3 k}{2}, b \leqslant \frac{n}{3}$.
Thus, $k\frac{n}{3} \geqslant b>a$.
So $n \mid(n-k)!$.
(ii) $a=b$, then $n=a^{2}, n-k>\frac{n}{3}=\frac{a^{2}}{3}$,

Since $k \geqslant 14$, then $p_{k} \geqslant 13, n \geqslant 26, a \geqslant 6$,
Thus $\frac{a^{2}}{3} \geqslant 2 a$. Hence $n-k>2 a$. So, $n \mid(n-k)!$.
If $a=2$, since $n \geqslant 26$, assume $b$ is not a prime, then $b=b_{1} b_{2}\left(b_{1} \leqslant b_{2}\right)$, since $b \geqslant 13$, then $b_{2} \geqslant 4$, thus $a b_{1} \geqslant 4$ falls into the case of $a \geqslant 3$,
Assume $b$ is a prime, then $b=\frac{n}{2}>p_{k}$,
Since $p_{k}$ is the largest prime less than $k$, then $b>k$,
Thus $n-k=2 b-k>b$. So, $n \mid(n-k)!$.
In summary, when $n>2 p_{k}$, $n \mid(n-k)!$.