Finding a formula for $f^{(n)} (x)$ if $f(x) = \ln(x-1)$ While working through Stewart's Calculus Late Transcendental 7th edition, I came across this problem:

Find a formula for $f^{(n)} (x)$ if $f(x) = \ln(x-1)$.

Obviously, I calculated the first few derivatives to see if I could spot a pattern:
$$f^{1}(x) = \frac1 {x-1}$$
$$f^{2}(x) = \frac{-1} {(x-1)^2}$$
$$f^{3}(x) = \frac{2}{(x-1)^3}$$
$$f^{4}(x) = \frac{-6}{(x-1)^4}$$
$$f^{5}(x) = \frac{24}{(x-1)^5}$$
I see the pattern of how the numerator is getting multiplied by $-1$ from the first to second derivative, $-2$ from $f^{2}(x)$ to $f^{3}(x)$, then $-3$ from $f^{3}(x)$ to $f^{4}(x)$, etc while the denominator simply increases by one power for each $nth$ derivative. 
Despite seeing a pattern, I am at a loss of expressing this as a formula. The answer given by Stewart is:
$$f^{n}(x) = \frac{(-1)^{n-1} (n-1)!} {(x-1)^{n}}$$
Any advice on how to  come up with a general formula when you see a pattern, but the formula isn't quite so obvious as with the one above?

You got really far in expressing your answer; this is essentially what you came up with from your textual explanation if you simply separate the negative (breaking down what you've figured out into its parts is a good strategy):
$$f^n(x)=\frac{(-1)^{n-1}\prod\limits_{k=1}^{n-1} k}{(x-1)^n}$$
If you put it in a notation like this, you can easily recognize that $\prod\limits_{k=1}^{n-1} k$ is simply a $(n-1)!$ factorial function.
Often times writing down what you figured out (even if it's a series) to visualize it mathematically is a good way to determine a more generalized formula by helping you recognize ways to simplify it.
So, in short:


*

*Separate everything you've figured out into parts and attempt to write an equation that uses each of those individual parts to make a whole.

*Even if one of the individual parts seems silly (like a series), write it out anyway.

*Simplify your result based on what you come up with.

