As to the original question, primes of the form $n!\pm 1$ are known as factorial primes and not all are known. It is in general a complicated question to determine if a number is prime or not, and only partial results are known. For example, if $n+1$ is prime then $n!+1$ is not.
As for the exercise which prompted this question, proving that there exists some $n$ such that $n,n+1,n+2,\dots,n+200$ are all composite consider the following:
Suppose we want to force each $n+i$ to be composite. If we want to force $2\mid n$ and $3\mid (n+1)$ and $5\mid (n+2)$, etc... that would correspond to the system of congruencies:
$\begin{cases} n\equiv 0\pmod{2}\\
n+1\equiv 0\pmod{3}\\
n+2\equiv 0\pmod{5}\\
\vdots\\
n+200\equiv 0\pmod{p_{201}}\end{cases}$
Consider then the Chinese Remainder Theorem.
The Chinese remainder theorem states that we can find such an $n$ that satisfies all of those congruencies since each of what we are modding out by are relatively prime to one another in every case.
Note: there is nothing intrinsically special about ordering these as being modulo $2$ followed by $3$ followed by $5$, etc... So long as we pick a list of length 200 where each of the entries on the list are coprime to one another, this will work.
Edit: Minor missing detail. It is possible that $n+i=p_i$ in one of those cases. To account for this possibility, technically chinese remainder will give us a solution to $n\equiv k\pmod{\prod p_i}$, so we can avoid this by instead of taking the smallest positive integer $n$ that works, by instead taking $n+\prod p_i$.
