[proofplan] We prove the statement by applying the [strictness of the Turing jump](/theorems/5417) to a representative of the degree $0^{(n)}$. The representative is the set $\varnothing^{(n)}$, whose jump is exactly $\varnothing^{(n+1)}$ by the recursive definition of the finite iterated jumps. Passing from sets to degrees then gives the strict inequality of Turing degrees. [/proofplan] [step:Choose the canonical representative of the degree $0^{(n)}$] Fix $n \in \mathbb{N} \cup \{0\}$. Define the set \begin{align*} A_n := \varnothing^{(n)} \subseteq \mathbb{N}. \end{align*} By the definition of $0^{(n)}$, the Turing degree of $A_n$ is $0^{(n)}$: \begin{align*} \deg_T(A_n) = 0^{(n)}. \end{align*} [/step] [step:Apply strictness of the Turing jump to the representative] The strictness theorem for the Turing jump states that for every set $A \subseteq \mathbb{N}$, \begin{align*} A <_T A'. \end{align*} (citing a result not yet in the wiki: Strictness of the Turing Jump) Applying this theorem to the set $A_n = \varnothing^{(n)}$, we obtain \begin{align*} \varnothing^{(n)} <_T (\varnothing^{(n)})'. \end{align*} [guided] We need a strict comparison between the $n$-th and $(n+1)$-st finite jumps. The general tool is the strictness theorem for the Turing jump: for every set $A \subseteq \mathbb{N}$, the jump $A'$ computes $A$, but $A$ does not compute $A'$. In symbols, \begin{align*} A <_T A'. \end{align*} (citing a result not yet in the wiki: Strictness of the Turing Jump) The hypothesis of this theorem is only that $A$ is a subset of $\mathbb{N}$. Our chosen representative \begin{align*} A_n := \varnothing^{(n)} \end{align*} is a subset of $\mathbb{N}$ by the recursive construction of the iterated Turing jumps. Therefore the theorem applies with $A = A_n$, and gives \begin{align*} A_n <_T A_n'. \end{align*} Substituting back $A_n = \varnothing^{(n)}$, this is \begin{align*} \varnothing^{(n)} <_T (\varnothing^{(n)})'. \end{align*} This is the entire source of strictness in the proof. [/guided] [/step] [step:Identify the jump of the representative with the next finite jump] By the recursive definition of the iterated Turing jumps, \begin{align*} \varnothing^{(n+1)} = (\varnothing^{(n)})'. \end{align*} Substituting this identity into the strict reducibility obtained above gives \begin{align*} \varnothing^{(n)} <_T \varnothing^{(n+1)}. \end{align*} [/step] [step:Pass from representatives to Turing degrees] Since strict Turing reducibility between representatives induces strict ordering of their Turing degrees, the relation \begin{align*} \varnothing^{(n)} <_T \varnothing^{(n+1)} \end{align*} implies \begin{align*} \deg_T(\varnothing^{(n)}) <_T \deg_T(\varnothing^{(n+1)}). \end{align*} Using the definitions \begin{align*} \deg_T(\varnothing^{(n)}) = 0^{(n)} \end{align*} and \begin{align*} \deg_T(\varnothing^{(n+1)}) = 0^{(n+1)}, \end{align*} we conclude that \begin{align*} 0^{(n)} <_T 0^{(n+1)}. \end{align*} Because $n \in \mathbb{N} \cup \{0\}$ was arbitrary, the strict inequality holds for every finite level $n$. [/step]