real analysis - Evaluating $lim_{n to infty} nleft(1-frac{1}{e}left(1+frac{1}{n}right)^{n} right)$

$$\lim_{n \to \infty} n\bigg(1-\dfrac{1}{e}\bigg(1+\dfrac{1}{n}\bigg)^{n} \bigg)$$

If I write expansion of $\bigg(1+\dfrac{1}{n}\bigg)^{n}$ it was equal to expansion of $e$ so $n-n=0$. Is limit is zero ?

Edit:
Uploading screen shot of my response. They marked it correct and I don't know if answer key is wrong or not ? Please can anyone say for sure like with $100$ percent surety if answer given is wrong.

Answer

By the L'Hôpital's rule we obtain:
$$\lim_{n\rightarrow+\infty}n\left(1-\frac{1}{e}\left(1+\frac{1}{n}\right)^{n}\right)=\lim_{x\rightarrow0^+}\frac{e-(1+x)^{\frac{1}{x}}}{ex}=-\frac{1}{e}\lim_{x\rightarrow0^+}\left(e^{\frac{\ln(1+x)}{x}}\right)'=$$
$$=-\frac{1}{e}\lim_{x\rightarrow0^+}(1+x)^{\frac{1}{x}}\left(\frac{1}{x(1+x)}-\frac{\ln(1+x)}{x^2}\right)=$$
$$=-\frac{1}{e}\lim_{x\rightarrow0^+}(1+x)^{\frac{1}{x}}\lim_{x\rightarrow0^+}\frac{x-(1+x)\ln(1+x)}{x^2+x^3}=$$
$$=-\lim_{x\rightarrow0^+}\frac{1-\ln(1+x)-1}{2x+3x^2}=\lim_{x\rightarrow0^+}\frac{\ln(1+x)}{x}\lim_{x\rightarrow0^+}\frac{1}{2+3x}=\frac{1}{2}.$$