My statistics textbook use below identity for find the value of n-th cumulant:
for $-1\lt t\lt 1$, $-\log(1-t)=- \sum_{n=1}^\infty{(-1)^{n+1} \over n}(-t)^n =\sum_{n=1}^\infty {(n-1)!\over n!}t^n$
However, how could one know the first identity of the above equality?
It comes from the power series expansion of $\log$ about $x=1$:
$$\log(1+x)=\sum_{n\geq 1}\,\frac{{(-1)}^{n+1}}{n}\,x^n,$$
which is convergent for $|x|<1$.
Taylor series expansion of $f(x) = \log(1-x)$ about $a = 0$ is,
$$f(x) = f(0) + \frac{f^{(1)}(a)(x-a)}{1!} + \frac{f^{(2)}(a)(x-a)^2}{2!} + \ldots $$
$$\implies \log(1-x) = \log(1) + \frac{\frac{-1}{1-x}\bigg\vert_{x=0}x}{1!} + \frac{\frac{-1}{(1-x)^2}\bigg\vert_{x=0}x^2}{2!} + \ldots + \frac{\frac{-(k-1)!}{(1-x)^k}\bigg\vert_{x=0}x^k}{k!}+\ldots$$
$$\implies \log(1-x) = -\frac{x}{1!} - \frac{x^2}{2!} - \ldots - \frac{x^k}{k}+\ldots$$
Note that in your equality, $(-1)^{n+1}(-t)^n = (-1)^{2n+1}t^n = -t^n$.
There is another way to derive the equality. For $|x|<1$,
$$\frac{1}{1-x} = 1+x+x^2+x^3+\ldots$$
Integrate both sides with respect to $x$, we get,
$-\log(1-x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{4} + \ldots$
