Now, what familiar equations can I use?

The Maclaurin series for $-\ln(1 - x)$ is $\displaystyle\sum_{n=1}^{\infty} \dfrac{x^{n}}{n}$ for $|x| \lt 1.$

In this case we have $x = 1 - \dfrac{1}{e},$ and thus the given series has the value

$-\ln\left(1 - \left(1 - \dfrac{1}{e}\right)\right) = -\ln\left(\dfrac{1}{e}\right) = -\ln(e^{-1}) = \boxed{1}.$

Proof of series expansion: Note first that the derivative of $-\ln(1 - x)$ is $\dfrac{1}{1 - x}.$

So for $|x| \lt 1$ we have that $-\ln(1 - x) = \displaystyle\int \dfrac{1}{1 - x} dx =$

$\displaystyle\int \left(\sum_{n=0}^{\infty} x^{n} \right) dx = \displaystyle\sum_{n=0}^{\infty} \left(\int x^{n} dx \right) = \sum_{n=0}^{\infty} \dfrac{x^{n+1}}{n + 1} + C = \sum_{n=1}^{\infty} \dfrac{x^{n}}{n + 1} + C.$

Now since $\ln(1) = 0$ we see that $C = 0,$ resulting in the aforementioned series representation.

Comments: For a discussion on the permissibility of exchanging summation and integral signs, see this link .

In general, for $y \ge 1$ we have that $\displaystyle\sum_{n=1}^{\infty} \dfrac{(1 - \frac{1}{y})^{n}}{n} = \ln(y).$