Prime gaps

Construction. Let $N=(n+1)!$. Then for $k=2,\ldots,n+1$:
$$N+k=k\cdot\frac{(n+1)!}{k}+k=k\left(\frac{(n+1)!}{k}+1\right).$$
Since $k$ divides $(n+1)!$ and $k\ge2$, $N+k$ is composite.

Count. There are $n$ consecutive composite numbers $N+2,\ldots,N+n+1$.

Conclusion. The gaps between consecutive primes can be arbitrarily large.