For what $n$ can the following system of inequalities be solved?

$$
1<x<2 ; \quad 2<x^{2}<3 ; \quad \ldots, \quad n<x^{n}<n+1
$$
Let $n \geq 3$, and suppose that the system of inequalities is solvable. Then

$$
\begin{gathered}
12(n-2) < 2(n-2)(n+1). \text{ On the other hand, by (4) } x^{n}<n+1, \text{ so } 2(n-2)<n+1, \text{ from which } n<5.
\end{gathered}
$$

We have shown that if $n \geq 5$, then the system of inequalities has no solution. We will show that if $n=1,2,3$ or 4, then it is solvable. It is sufficient to prove this for $n=4$, which precisely means that the common part of the following four intervals is not empty:

$$
I_{1}=(1 ; 2), \quad I_{2}=(\sqrt{2} ; \sqrt{3}), \quad I_{3}=(\sqrt[3]{3} ; \sqrt[3]{4}), \quad I_{4}=(\sqrt[4]{4} ; \sqrt[4]{5})
$$

Since $1<\sqrt{2}<\sqrt{3}<2$, it follows that $I_{2} \subset I_{1}$. From the sequence of inequalities $2^{3}<3^{2}<4^{2}<3^{3}$, taking the sixth root, we get that $I_{3} \subset I_{2}$. Our claim now follows from the fact that the intersection of $I_{3}$ and $I_{4}$ is not empty. From the chain of inequalities $4^{3}<3^{4}<5^{3}<4^{4}$, taking the 12th root,

$$
\sqrt[4]{4}<\sqrt[3]{3}<\sqrt[4]{5}<\sqrt[3]{4}
$$

so $I_{3} \cap I_{4}=(\sqrt[3]{3} ; \sqrt[4]{5})$, and for example, 1.45 is an element of this interval.

Thus, the system of inequalities is solvable for $n=1,2,3$, and 4.

Remark. The fact that there is no solution for $n \geq 5$ also follows from the fact that the system of inequalities $3<x^{3}<4$ and $5<x^{5}<6$ has no solution, because

$$
\sqrt[5]{6}<\sqrt[3]{3} \cdot\left(6^{3}=216<243=3^{5}\right)
$$