Does $\sum_{n=0}^\infty (-1)^n (e-(1+\frac{1}{n})^n)$ converge absolutely, conditionally, or diverge? I've been given the hint to use the binomial theorem and show that $e-\left(1+\frac{1}{n}\right)^n > \frac{1}{2n}$ for $n \geq 2$. So I've written
\begin{align*}
e-\left(1+\frac{1}{n}\right)^n = e - \sum_{k=0}^n {n \choose k} \frac{1}{n^k}
\end{align*}
But I'm not sure where to go from here to assess convergence. What are the next steps?

Converge absolutely? WRONG. Because as you said, $e-\left(1+\frac{1}{n}\right)^n > \frac{1}{2n}$.
Let us prove this inequality.
\begin{align*}e-\left(1+\frac{1}{n}\right)^n>{}&\left(1+\frac{1}{2n}\right)^{2n}-\left(1+\frac{1}{n}\right)^n \\
={}&\left(1+\frac{1}{n}+\frac{1}{4n^2}\right)^{n}-\left(1+\frac{1}{n}\right)^n\\
={}&\left(\left(1+\frac{1}{n}+\frac{1}{4n^2}\right) -\left(1+\frac{1}{n}\right)\right)\sum_{i=0}^{n-1}\left(1+\frac{1}{n}+\frac{1}{4n^2}\right)^i\left(1+\frac{1}{n}\right)^{n-1-i}\\
>{}&\left(\left(1+\frac{1}{n}+\frac{1}{4n^2}\right) -\left(1+\frac{1}{n}\right)\right)\left(1+\frac{1}{n}\right)^{n-1}n\\
={}&\frac{1}{4n}\left(1+\frac{1}{n}\right)^{n-1}\\
>{}& \frac{1}{2n}
\end{align*}
The last inequality holds for $n$ large enough.
Converge conditionally? YES. Since $e-\left(1+\frac{1}{n}\right)^n$ decreases monotonically and goes to $0$. (Leibniz criterion)
