We provide three methods to show that $c_n = n \cdot (n+1)!/2 - 1$ for all $n \in \mathbb{N}$.
(We omit the assumption $n > 1$ when proving this.) Let $d_n := c_n + 1$, i.e. $d_n$
is the number of $(n+1)$-step methods including that of the teacher.

Method 1. Consider the first step of any $(n + 1)$-step method. If this involves moving some switch $s_1$ to position 2, we call it a Type A step. After we perform a Type A step on $s_1$, we cannot subsequently move $s_1$ since we would later need to move it back again to position 2, thereby “wasting” two steps when we only have a single step to spare. Our only option therefore is to use the remaining $n$ steps to change the other $n-1$ switches from position 1 to position 2. This can be done in $d_{n-1}$ ways. Since there are $n$ choices for the initial switch $s_1$, we get a total of $nd_{n-1}$ methods that start with a Type A step.
Alternatively, if the first step moves a switch to position 3, we call it a Type B step. After a Type B step, all $n$ switches are in the wrong position, so each subsequent step must consist of changing a switch to position 2. Thus, the only choices involve the order in which we carry out these moves, giving a total of $n!$ methods that start with the given Type B step. Since there are $n$ choices for the initial switch $s_1$, there are $n \cdot n!$ methods that start with a Type B step.

Rewriting these equations in terms of $e_n := d_n/n!$, we get $e_n = e_{n-1} + n$ for
all $n > 1$, and $e_1 = 1$. Thus, $e_n = n(n+1)/2$ and $d_n = n \cdot (n+1)!/2$. This
finishes the proof of our claim.

Method 2. First note that in each $(n+1)$-step method, exactly one switch will first be switched to 3 and then to 2, whereas all others have to be switched to 2 directly. If we ignore this extra initial switching to 3, each $(n+1)$-step method gives rise to an $n$-step method, which is just an ordering of the switches; and there are $n!$ of them.

For how many different $(n + 1)$-step methods do we get to the same $n$-step
method (or ordering of switches)? The answer is $1+2+3+...+n$, because the
switch which is the $k$-th which is switched to 2 in the $n$-step method, could be switched to 3 any time before it is switched to 2, and there are $k$ opportunities to do so. The conclusion is that $d_n = n! \cdot n \cdot (n+1)/2 = n \cdot (n+1)!/2$.

Method 3. Exactly one switch has to be touched twice. If this is switch $k$, then we are seeking the number of ways to permute the numbers
$$1, 2, 3, \dots, k-1, k, k, k+1, \dots, n,$$
which is $(n+1)!/2$. Taking into account all $n$ values of $k$ yields a total of $d_n = n \cdot (n+1)!/2$.

We now prove the divisibility statements about $c_n = n \cdot (n+1)!/2 - 1$. Since $(n+1)!$ is even for all $n > 0$, $c_n$ is an integer. Since $n > 1$ by assumption, $c_n > 0$. (Of course, this is also clear in terms of the combinatorial meaning.) To show that $c_n$ is not divisible by any integer $1 < k < n + 2$, it suffices to prove this under the additional assumption that $k$ is prime.
If $n$ is even, $c_n = \frac{n}{2} \cdot (n+1)! - 1 \equiv -1 \pmod{k}$ for any integer $1 < k < n+2$. If $n$ is odd, $n+1 > 2$ is even hence not prime. Thus, any prime $k$ that satisfies $1 < k < n + 2$, is a factor of $n!$ and so cannot divide $c_n = \frac{n+1}{2} \cdot n \cdot n! - 1$.

It remains to consider divisibility by $n+2$. If $n+2$ is not prime, then it has a prime factor $k$ less than $n+2$. By the above, $c_n$ is not divisible by $k$, and so it is not divisible by $n+2$.
Suppose finally that $p := n + 2 > 3$ is prime. By Wilson's theorem,
$$(n + 1)! \equiv -1 \pmod{p}.$$
Since also $n \equiv -2 \pmod{p}$, the equation $2d_n = n(n+1)!$ implies that
$$2d_n \equiv (-2)(-1) \equiv 2 \pmod{p},$$
and so $d_n \equiv 1 \pmod{p}$, because $\gcd(2, p) = 1$. Thus, $p = n + 2$ divides $c_n = d_n - 1$ as required.