How did they get this result starting in ($\star$)?
Define:
$X_0 := 1, X_{n+1} = \sum_{i=1}^{X_n} Y_i$
with
$Y_i$ iid for all $i = 1,2,\ldots$, and $\mathbb{E}(Y_i) = \mu, \operatorname{Var}(Y_i) = \sigma^2$.
We found that $\mathbb{E}(X_n) = \mu^{n}$.
We want to find the variance of $X_{n+1}$.
$$\operatorname{Var}(X_{n+1})= \mathbb{E}(\operatorname{Var}(X_n \mid X_{n+1}) + \operatorname{Var}(\mathbb{E}(X_{n}\mid X_{n+1}) \\ = \sigma^2 \mu^{n-1} + \mu^2 \operatorname{Var}(X_{n-1})$$
Since $\operatorname{Var}(X_0) = 0$ and $\operatorname{Var}(X_1) = \sigma^2$, we obtain:
$$\operatorname{Var}(X_n) = \sigma^2 (\mu^{n-1} + \mu^n + \cdots +\mu^{2n-2}) \qquad (\star)$$
The formula before $(\star)$ is incorrect. It should read: $$ \begin{align} \operatorname{Var}(X_{n})&= \mathbb{E}(\operatorname{Var}(X_n \mid X_{n-1})) + \operatorname{Var}(\mathbb{E}(X_n \mid X_{n-1})) \\ &= \sigma^2 \mu^{n-1} + \mu^2 \operatorname{Var}(X_{n-1}) \end{align} $$ You should be able to get to $(\star)$ from here.