Let's determine for every $n$ all $n$-digit (in decimal system) numbers that are equal to the $n$-th power of the sum of their digits.
Let's denote the sum of the digits by $S$. Since $S^{n}$ is an $n$-digit number, we have

$$
10^{n-1} \leq S^{n}<10^{n}
$$

and thus

$$
\frac{10}{\sqrt[n]{10}} \leq S<10
$$

For the case $n=1$, clearly every single-digit number satisfies the requirements.

If $n \geq 2$, then according to (1), $S \geq 10 / \sqrt{10}$, so $S \geq 4$.

The numbers $S=4,5,6$ do not have any powers that would meet the criteria. Among the powers of $4$, only $4^{2}$ has a number of digits equal to the exponent, but the sum of its digits is not equal to the base. In the powers of $5^n$ and $6^n$, the last digit is always 5 and 6, respectively, so the sum of the digits (for $n \geq 2$) is always greater than the base.

For the base $S=7$, the first and only suitable power is the fourth: $7^{4}=2401$, the sum of its digits is 7. The subsequent powers $7^{5}$ and $7^{6}$ already have a last digit that is too large (7 and 9), and $7^{7}$ is already only six digits long.

For the sum $S=8$, the only suitable power is the cube: $8^{3}=512$. The difference $S^{n}-S$ is divisible by 9 if $S^{n}$ is a suitable number, since $S^{n}$ gives the same remainder when divided by 9 as the sum of its digits, which is exactly $S$ for a suitable number. Therefore, only those exponents $n$ are worth examining for $S=8$ where $\left(S^{n-1}-1\right)$ is divisible by 9. The powers of 8, when divided by 9, alternately give remainders of 8 and 1, so only odd exponents can be considered. The last two digits of these are:

$$
8^{3}=\ldots 12, \quad 8^{5}=\ldots 68, \quad 8^{7}=\ldots 52, \quad 8^{9}=\ldots 28, \quad 8^{11}=\ldots 92
$$

Among these, $8^{3}$, as we have seen, is suitable, $8^{5}$ and $8^{9}$ already have a last digit that is too large, $8^{7}=\left(2^{10}\right)^{2} \cdot 2$ has a first digit of 2, so the last two digits 5 and 2 are too large, and finally $8^{11}<10^{10}$, and from this point on, the number of digits in the powers of 8 is already less than the exponent.

Finally, if $S=9$, then $S^{2}=81$ is the only suitable power. Taking the logarithm of (1), we get

$$
1-\frac{1}{n} \leq \lg 9
$$

so $n \leq 1(1-\lg 9)$, i.e., $n \leq 21$. Since the powers of 9 alternately end in 9 and 1, only the powers ending in 1, i.e., those with even exponents, can be considered. We will calculate the last three digits of the first 10 powers of $81$. Let $A_{n}$ denote the hundreds digit and $B_{n}$ the tens digit in $81^{n}$, then

$$
\left(100 A_{n}+10 B_{n}+1\right)(80+1)=800 A_{n}+\left(10 A_{n}+B_{n}+8\right) 10+1
$$

Accordingly, we need to take the last two digits of the number $10 A_{n}+B_{n}+8+80 A_{n}$:

$$
\begin{gathered}
81^{2}=\ldots 561, \quad 81^{3}=\ldots 441, \quad 81^{4}=\ldots 721, \quad 81^{6}=\ldots 401, \quad 81^{6}=\ldots 481 \\
81^{7}=\ldots 961, \quad 81^{8}=\ldots 841, \quad 81^{9}=\ldots 121, \quad 81^{10}=\ldots 801
\end{gathered}
$$

So only $81^{5}$ and $81^{9}$ can be considered. We will calculate the last four digits of these:

$$
81^{2}=\ldots 6561, \quad 81^{4}=\ldots 6721, \quad 81^{8}=\ldots 1841, \quad 81^{5}=\ldots 4401, \quad 81^{9}=\ldots 9121
$$

Thus, these do not meet the criteria either.

This completes our examination, essentially finding the following numbers to be suitable: for $n=1$: $1,2,3,4,5,6$, $7,8,9$; for $n=2$: $81$; for $n=3$: $512$; for $n=4$: $2401$.