Let $(R,+,·)$ be a ring with unity. If $e$ is idempotent, prove that $1-e$ is also idempotent.
Here's my attempt: If $e$ is idempotent then $e^n=e$. Then $$e=e^n$$ $$e-e^n=0$$ $$e(1-e^{n-1})=0$$ $$1-e^{n-1}=0$$ But $e^{n-1}=e$, so: $$1-e=0$$ which means $1-e$ is idempotent. Is this correct?
I will assume that throughout your question, $n = 2$.
No, that is not correct, because given that $e (1 - e) = 0$, we cannot conclude that either $e = 0$ or $1 - e = 0$. It's possible for both $e$ and $1 - e$ to be nonzero, and yet for $e (1 - e)$ to be $0$ nevertheless.
For example, in the ring of $2 \times 2$ matrices (where $0$ denotes the zero matrix and $1$ denotes the identity matrix),
$$1 - \begin{bmatrix} 1 & 0 \\ 0 & 0\end {bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 1\end {bmatrix} \ne 0,$$
and yet
$$\begin{bmatrix} 1 & 0 \\ 0 & 0\end {bmatrix} \left (1 - \begin{bmatrix} 1 & 0 \\ 0 & 0\end {bmatrix} \right) = \begin{bmatrix} 1 & 0 \\ 0 & 0\end {bmatrix} \begin{bmatrix} 0 & 0 \\ 0 & 1\end {bmatrix} = 0.$$
However, in any ring, $(1 - e)^2 = (1 - e)(1 - e) = 1 - 2 e + e^2$. Since $e^2 = e$, what can you conclude from this?