Deterministic Parity Automata require unbounded index Deterministic parity automata $(Q, \Sigma, q_0, \Delta, c)$ are powerful enough to recognize all $\omega$-regular languages. However, the number of priorities they require for a language can become arbitrarily high. For example, the languages $L_{i,j} = \{ \alpha \in \{i, \dots, j\}^\omega \mid \text{The highest number occurring infinitely often in } \alpha \text{ is even} \}$ require at least $j-i+1$ priorities. I was not able to prove this theorem so far by myself, in particular I am stuck at the case that the automaton uses exactly the priorities from $i+1$ to $j$. I could not lead this assumption to a contradiction.

Any help or link to a paper would be appreciated.
Take a look at the following paper:
Relating hierarchies of word and tree automata.
They cite the result you are interested in.

As a more concrete answer, consider the following argument: let's look at the language $M_n=\{w\in \{0,...,n\}^\omega: \limsup w_i \text{ is even}\}$ (the same as your language with $i=0$), and assume it has a parity automaton $A$ with ranks $1,...,n+1$ (note that proving it doesn't have a parity automaton with ranks $1,...,n$ is not enough, as it's still possible for it to have ranks $0,...,n-1$).

We will construct a word that is accepted by $A$, but is not in the language (or vice-versa). Denote by $k$ the number of states in $A$. Consider the run of $A$ on the word $0^\omega$. It's accepting, so the limsup degree is even, and so at least $2$. Moreover, it must occur during the first $k$ steps, otherwise, since $A$ is deterministic, we have a non-accepting cycle. In addition, this is true for every $0^k$ infix of any word.

Now, consider the word $(0^k1)^\omega$. This word is not accepted, so the limsup degree is odd, but must be at least 3, since the $0^k$ infixes induce the degree $2$. Finally, this odd degree must occur within the first $k$ cycles of $0^k1$, for similar reasons as above.

We can repeat these alternating examples, constructing a word of the form $(0^k1)^k2$, and so on, with each word requiring a higher degree. However, since we only have degrees $1,...,n+1$, then eventually we will have a word that uses the letters $0,...,n-1$, whose limsup degree is $n+1$. From there, continuing this construction will still leave the limsup degree $n+1$, but will change the word from accepted to non-accepted, or vice-versa. In either case, the automaton fails to recognize the language.