Why is the reduction-formula for $\int\sec^n(x)dx$ only valid for $n\ge3$? I have a question regarding the following reduction formula:
$$I_n=\int\sec^n(x)dx=\frac{1}{n-1}(\sec^{n-2}x\tan x)+\frac{n-2}{n-1}I_{n-2}+C$$
My calculus book states that it is only valid for $n\ge3$. Why is this the case? How does one intuit such a result?
Sure enough, $I_1$ breaks down because of division by zero. But what about $n=2$? And why can't $n=0$ be a base-case? Or even $n=-1$ etc.?
Finally: How do I know for sure it will actually work with all $n\ge3$? 

Consider the beginning of the derivation:
$$\int \sec^n(x) dx = \int \sec(x)^{n-2} \sec(x)^2 dx = \sec(x)^{n-2} \tan(x) - \int (n-2) \sec(x)^{n-2} \tan(x)^2 dx.$$
In the case $n=1$, this is true but it is not useful, because you get the same multiple of $I_1$ on the right side as you already had, so you can't isolate $I_1$. You just get the trivial equation $I_1=I_1$.
In the case $n=2$, this technically works (the second term is just zero). But it only works because you've already used the answer in taking the first step (you had to integrate $\sec(x)^2$ to do the integration by parts in the first place). So it doesn't make sense to think of it as part of the recursion for even $n$, instead it is the base case of the recursion for even $n$.
