Show that $nY_n$ converges in distribution

Let $X_1,...$ be a sample from the distribution $$f(x)=\frac{1}{2}(1+x)e^{-x}.$$

Set $Y_n=\min\left\{X_1,...,X_n\right\}$. Show that $nY_n$ converges in distribution as $n$ tends to infinity.

So far I have $$F(x)=1-\frac{1}{2}e^{-x}(x+2).$$

$$P(nY\geq y)= P(Y\geq \frac{y}{n})=P(\min(X_{1},X_{2},..,X_{n})\geq \frac{y}{n})=\\P(X_{1}\geq \frac{y}{n})P(X_{2}\geq\frac{y}{n})...P(X_{n}\geq\frac{y}{n})=P(X_{1}\geq\frac{y}{n})^{n}=\\(1-P(X_{1}\leq\frac{y}{n}))^{n}=\frac{1}{2^{n}}e^{-\frac{y}{n}\cdot n}(2+\frac{y}{n})^{n}=e^{-y}(1+\frac{y}{2n})^{n}$$.

Thus $$\lim_{n\to\infty} P(nY\geq y)=\lim_{n\to\infty}e^{-y}(1+\frac{y}{2n})^{n}=e^{-\frac{y}{2}}$$

So $F_{nY}(y)=P(nY\leq y)\to1-e^{-\frac{y}{2}},\,y\geq 0$.

Which is the same as the cdf of a $\text{Exp}(\frac{1}{2})$ variate.

Thus $nY\xrightarrow{d}Z$.

Where $Z\sim \text{Exp}(\frac{1}{2})$