The formula for permutations is ^kP_m = (k!)/((k-m)!).

Since $n!$ is on both sides and $n \ge 4$ (for $^nP_4$ to be defined), $n! \neq 0$. We can divide both sides by $n!$.

We know that $(n-2)! = (n-2)(n-3)(n-4)!$.

We look for two numbers that multiply to -14 and add to -5. These numbers are -7 and 2.

This gives two possible solutions:

For $^nP_r$ to be defined, $n$ must be a non-negative integer and $n \ge r$. In this case, $n \ge 4$.
The value $n=7$ satisfies $n \ge 4$.
The value $n=-2$ does not satisfy $n \ge 4$.

Therefore, the only valid solution is $n=7$.

The value of $n$ is $\boxed{\text{7}}$.