Proving a set is at a certain level of the arithmetical hierarchy
In order to show that a language is in (say) $\Sigma_n^0$ but not in $\Sigma_{n-1}^0$ or $\Pi_{n-1}^0$, we show that the language is complete for $\Sigma_n^0$. Since the arithmetical hierarchy is strict (by a generalization of the proof of the uncomputability of the halting problem by diagonalization, which you can find in many different sources), a $\Sigma_n^0$-complete language cannot be in any lower level of the hierarchy.

How do we show that a language is complete for a level of the hierarchy? Either by a direct proof or by reduction. Let me illustrate both techniques using variants of the halting problem.

We will consider two variants of the halting problem:


$H_1$ consists of all pairs $(\langle M \rangle, x)$ such that $M$ is a Turing machine halting on $x$.
$H_2$ consists of all singletons $\langle M \rangle$ such that $M$ is a Turing machine halting on an empty tape.


Let us show that $H_1$ is $\Sigma_1^0$-complete by a direct proof. In order to show that it is in $\Sigma_1^0$, we notice that $(\langle M \rangle,x)$ iff there exists $t$ such that $M$ halts on $x$ within $t$ steps; the latter predicate is computable.

In order to show that $H_1$ is $\Sigma_1^0$-hard, consider any language $L \in \Sigma_1^0$. By definition, there exists a computable predicate $P$ such that
$$ x \in L \Longleftrightarrow \exists y \, P(x,y). $$
Let $M$ be a Turing machine that on input $x$, enumerates all strings $y$ (say first by length, then lexicographically), for each one computes $P(x,y)$, and halts if $P(x,y)$ holds for any $y$. Then $x \in L$ iff $(\langle M \rangle, x) \in H_1$. The mapping $x \mapsto (\langle M \rangle, x)$ is clearly computable, and so this reduces $L$ to $H_1$.

We now show that $H_2$ is $\Sigma_1^0$-complete by reduction from $H_1$. Given an instance $(\langle M \rangle, x)$, we construct (in a computable way) a new machine $M_x$ that first initializes the tape $x$, and then transfers control to $M$. Clearly $(\langle M \rangle, x) \in H_1$ iff $\langle M_x \rangle \in H_2$. Since the reduction is computable and $H_1$ is $\Sigma_1^0$-hard, this shows that $H_2$ is also $\Sigma_1^0$-hard. Also, $H_2 \in \Sigma_1^0$, using the same proof as above. So $H_2$ is $\Sigma_1^0$-complete.