we will not assume this knowledge.

Since there is no copy of $H$ in $G_{n}$, it follows that $G_{n}[X_{n}]$ contains no copy of $H^{k}$. We have that $$\begin{aligned}
\pi(H) &=\lim_{n\to\infty} h_{n}(G_{n}) \\
             &=\lim_{n\to\infty} \sum_{F\in H^{1}}\frac{1}{\binom{n}{1}} + \sum_{F\in H^{k}} \frac{1}{\binom{n}{k}} \\
             &\leq \lim_{n\to\infty} \frac{x_{n}n}{\binom{n}{1}} + \frac{\binom{n}{k}-(1-\pi_{x_{n}n}(H^{k}))\binom{x_{n}n}{k}}{\binom{n}{k}} \\
         &=\lim_{n\to\infty} 1 + x_{n} - (1-\pi_{x_{n}n}(H^{k}))\frac{\binom{x_{n}n}{k}}{\binom{n}{k}} \\
         &\leq \lim_{n\to\infty} \begin{cases} 1+\frac{1}{\sqrt{n}} & \text{if } x_{n}n \leq \sqrt{n}, \\
                                                1+ x_{n} - (1-\pi_{x_{n}n}(H^{k}))x_{n}^{k} & \text{if } x_{n}n>\sqrt{n}, \end{cases} \\
         &\leq \max\{1, 1+x-(1-\pi(H^{k}))x^{k}\}.\end{aligned}$$ Let $f(x)=1+x-(1-\pi(H^{k}))x^{k}$ and then note that $$\pi(H)= \lim_{n\to\infty} h_{n}(G) \leq \max_{x\in [0,1]} f(x).$$ An easy calculus exercise shows that $f^{\prime\prime}(x)<0$ for all $x>0$, and $f^{\prime}(x)=0$ when $x=\left(\frac{1}{k(1-\pi(H^{k}))}\right)^{\frac{1}{k-1}}.$ If $\frac{1}{k(1-\pi(H^{k}))}\geq 1$ then $f^{\prime}(x)>0$ when $x\in [0,1)$ and hence $f(x)$ is maximized when $x=1$. Note that $f(1)=1+\pi(H^{k})$. If, on the other hand, $\frac{1}{k(1-\pi(H^{k}))}< 1$ it follows that $f(x)$ is maximized at $x=\left(\frac{1}{k(1-\pi(H^{k}))}\right)^{1/(k-1)}$. Together, this gives us $$\pi(H) \leq \begin{cases} 1+\pi(H^{k}) & \text{if } \pi(H^{k})\geq 1-\frac{1}{k}; \\
                            1+\left(\frac{1}{k(1-\pi(H^{k}))}\right)^{1/(k-1)}\left(1-\frac{1}{k}\right) & \text{otherwise}. \end{cases}$$ To get equality, take $x$ that maximizes $f(x)$ as above. For any $n\in \mathbb{N}$ (thinking of $n\to \infty$) partition $[n]$ into two sets $X$ and $\bar{X}$ with $|X|=xn$ and $|\bar{X}|=(1-x)n$. Let $E(G^{1})=\{\{v\}:v\in X\}$ and let $g^{k}$ be a $k$-uniform graph on $xn$ vertices attaining $|E(g^{k})|=\text{ex}(xn,H^{k})$ 