I don't see where the $\frac{1}{(n+1)!}$ factor comes from.

If you take $II = \int_{a}^{b}\frac{1}{t^{n+2}}dt$ and perform the change of variables $t \to 1/x$, you get
$$-\int_{1/a}^{1/b}\left(\frac{1}{x}\right)^{-n}dx$$
Taking $a>0$, $b>0$ lets one flip things around. As $x>0$, $\left(\frac{1}{x}\right)^{-n}$ is equal to $x^n$. Letting $a\to 0^{+}$ and $b\to\infty$, you get
$$II = \int_{0}^{\infty} x^{n} dx$$
Purely formally, of course. But it is hard to see how that factor would 'sneak in'.

The way it can sneak in is by the *process* you use to derive things. If you go via a larger space (like $\mathbb{C}$ say, or a larger space of function, or an integral transform), it is then entirely possible to have that process 'see' things that are invisible from the purely extensional question (i.e. the values of the functions on exactly and only the positive reals).

Roughly, it boils down to:

- for convergent integrals, we have all sorts of theorems that say that the answer is unique, i.e. no matter what computation process is used, the answer will always be the same.
- for divergent integrals, it is necessarily the case that the answer is process-dependent. There are still uniqueness theorems, but they all say "given class of process X, then the answer is unique".
- there is no guarantee that two classes of processes will agree on divergent integrals.

The very important part about point 3 above is: you can't mix processes, else you will get inconsistencies. All interpretations of all identities must be within the same framework, and results between frameworks can't trivially be ported.