We prove the requested statement by induction. For $n=1$, we only look at the number $1^{1}$, so the statement is trivially true.

Assume as the induction hypothesis that $1^{1}, 3^{3}, 5^{5}, \ldots,\left(2^{n}-1\right)^{2^{n}-1}$ are in different residue classes modulo $2^{n}$. First, these numbers are also in different residue classes modulo $2^{n+1}$. Since $\varphi\left(2^{n+1}\right)=2^{n}$, it holds that $a^{k} \equiv a^{\ell} \bmod 2^{n+1}$ if $k \equiv \ell \bmod 2^{n}$ and $a$ is odd.

The additional numbers we consider can be written as $\left(2^{n}+m\right)^{2^{n}+m}$ with $1 \leq m \leq 2^{n}-1$ and $m$ odd. If we expand $\left(2^{n}+m\right)^{2^{n}+m}$ using the binomial theorem, we notice that every term with at least two factors $2^{n}$ is congruent to 0 modulo $2^{n+1}$. Therefore, modulo $2^{n+1}$, we have

$$\begin{aligned} \left(2^{n}+m\right)^{2^{n}+m} & \equiv m^{2^{n}+m}+\left(2^{n}+m\right) m^{2^{n}+m-1} 2^{n}+\binom{2^{n}+m}{2} m^{2^{n}+m-2}\left(2^{n}\right)^{2}+\ldots \\ & \equiv m^{2^{n}+m}+\left(2^{n}+m\right) m^{2^{n}+m-1} 2^{n} \\ & \equiv m^{m}+\left(2^{2 n}+2^{n} m\right) m^{m-1} \\ & \equiv m^{m}+2^{n} \cdot m^{m} \\ & \equiv m^{m}+2^{n} \end{aligned}$$

where in the last step we used the fact that $m^{m}$ is odd. This means that we can write $m^{m}$ as $2a+1$, and then we find that $2^{n}(2a+1)=2^{n+1}a+2^{n} \equiv 2^{n} \bmod 2^{n+1}$.

Since the numbers $m^{m}$ from the first group are distinct modulo $2^{n+1}$, the numbers from the second group are also distinct modulo $2^{n+1}$. From this calculation, we also conclude that the numbers from the first group are different from the numbers from the second group. Suppose that $\left(2^{n}+m\right)^{2^{n}+m} \equiv k^{k} \bmod 2^{n+1}$ with $1 \leq k, m \leq 2^{n}-1$, then from this calculation it follows in particular that $m^{m} \equiv m^{m}+2^{n} \equiv k^{k} \bmod 2^{n}$. Therefore, by the induction hypothesis, we know that $m=k$. But in that case, it holds that $\left(2^{n}+m\right)^{2^{n}+m} \equiv m^{m}+2^{n} \not \equiv m^{m} \bmod 2^{n+1}$.

We conclude that no number from the groups $1^{1}, 3^{3}, 5^{5}, \ldots,\left(2^{n}-1\right)^{2^{n}-1}$ and $\left(2^{n}+1\right)^{2^{n}+1},\left(2^{n}+3\right)^{2^{n}+3}, \ldots,\left(2^{n+1}-1\right)^{2^{n+1}-1}$ has the same residue class as another number from these two groups. This completes the induction step, and by induction, the statement is true for all natural numbers $n$.