A limit of indeterminate

$$\begin{align}\lim_{x\to \infty}\frac{e^x}{x}((1+\frac{x}{e^x})^{n+1} - 1) &= \lim_{t\to \infty}t((1+\frac{1}{t})^{n+1}-1) \\&= \lim_{t\to \infty}t(1+\frac{n+1}{t}-1 + o(\frac{1}{t}))\\&=n+1\end{align}$$

Use binomial expansion on the numerator : $$\left(e^x+x\right)^{n+1}-e^{(n+1)x} = \sum_{k=0}^n \binom{k}{n+1} e^{kx}x^{n+1-k}$$ Now the dominant term here is the term of number $k=n$, so $$\frac{\left(e^x+x\right)^{n+1}-e^{(n+1)x}}{xe^{nx}} \sim \frac{(n+1)xe^{nx}}{xe^{nx}}=n+1$$ An alternative method, starting with your factorization, is to use $(1+u)^\alpha-1\sim \alpha u$, so : $$\frac{e^x}{x}\left(\left(1+\frac{x}{e^x}\right)^{n+1}-1\right) \sim \frac{e^x}{x}.(n+1)\frac{x}{e^x}$$ which gives the same result.