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Problem Overview:

The task involves solving a series of mathematical expressions involving indices (exponents) and basic arithmetic operations. Each solution must be a whole number between 1 and 24, and all numbers from 1 to 24 must appear exactly once.

Solution Approach:

We will solve each expression step by step, ensuring that the results are whole numbers between 1 and 24. We will also keep track of the numbers used to ensure no duplicates.

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Step-by-Step Solutions:

#### Column 1:
1. \( 7^5 \times \square^2 = 7^7 \)
- Using the index law \( a^m \times a^n = a^{m+n} \):
\[
7^5 \times 7^2 = 7^{5+2} = 7^7
\]
- Therefore, \( \square = 7 \).

2. \( 3^4 = \square^2 \)
- Calculate \( 3^4 \):
\[
3^4 = 81
\]
- Solve for \( \square \):
\[
\square^2 = 81 \implies \square = 9
\]

3. \( 2^2 \times 3 = \square \)
- Calculate:
\[
2^2 = 4 \quad \text{and} \quad 4 \times 3 = 12
\]
- Therefore, \( \square = 12 \).

4. \( 2^3 \times 2^5 = 2^\square \)
- Using the index law \( a^m \times a^n = a^{m+n} \):
\[
2^3 \times 2^5 = 2^{3+5} = 2^8
\]
- Therefore, \( \square = 8 \).

5. \( 2^3 \times 3 = \square \)
- Calculate:
\[
2^3 = 8 \quad \text{and} \quad 8 \times 3 = 24
\]
- Therefore, \( \square = 24 \).

6. \( \square^2 = 100 \)
- Solve for \( \square \):
\[
\square^2 = 100 \implies \square = 10
\]

7. \( 2^1 \times 11^1 = \square \)
- Calculate:
\[
2^1 = 2 \quad \text{and} \quad 11^1 = 11 \quad \implies 2 \times 11 = 22
\]
- Therefore, \( \square = 22 \).

8. \( 2^2 \times 5 = \square \)
- Calculate:
\[
2^2 = 4 \quad \text{and} \quad 4 \times 5 = 20
\]
- Therefore, \( \square = 20 \).

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#### Column 2:
9. \( 6^2 - 5^2 = \square \)
- Calculate:
\[
6^2 = 36 \quad \text{and} \quad 5^2 = 25 \quad \implies 36 - 25 = 11
\]
- Therefore, \( \square = 11 \).

10. \( 23^1 = \square \)
- Calculate:
\[
23^1 = 23
\]
- Therefore, \( \square = 23 \).

11. \( \square^2 = 361 \)
- Solve for \( \square \):
\[
\square^2 = 361 \implies \square = 19
\]

12. \( 3^7 \times 3^7 \times 3^7 = 3^\square \)
- Using the index law \( a^m \times a^n = a^{m+n} \):
\[
3^7 \times 3^7 \times 3^7 = 3^{7+7+7} = 3^{21}
\]
- Therefore, \( \square = 21 \).

13. \( 3 \times 3 \times 3 \times 3 = 3^\square \)
- Simplify:
\[
3 \times 3 \times 3 \times 3 = 3^4
\]
- Therefore, \( \square = 4 \).

14. \( 2^1 \times 3^1 = \square \)
- Calculate:
\[
2^1 = 2 \quad \text{and} \quad 3^1 = 3 \quad \implies 2 \times 3 = 6
\]
- Therefore, \( \square = 6 \).

15. \( \square^3 = 8 \)
- Solve for \( \square \):
\[
\square^3 = 8 \implies \square = 2
\]

16. \( \square^2 = 9 \)
- Solve for \( \square \):
\[
\square^2 = 9 \implies \square = 3
\]

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#### Column 3:
17. \( 2^4 + 1 = \square \)
- Calculate:
\[
2^4 = 16 \quad \text{and} \quad 16 + 1 = 17
\]
- Therefore, \( \square = 17 \).

18. \( 6^0 = \square \)
- Any non-zero number raised to the power of 0 is 1:
\[
6^0 = 1
\]
- Therefore, \( \square = 1 \).

19. \( 2 \times 3^2 = \square \)
- Calculate:
\[
3^2 = 9 \quad \text{and} \quad 2 \times 9 = 18
\]
- Therefore, \( \square = 18 \).

20. \( 2^3 \times 2^5 \times 2^7 = 2^\square \)
- Using the index law \( a^m \times a^n = a^{m+n} \):
\[
2^3 \times 2^5 \times 2^7 = 2^{3+5+7} = 2^{15}
\]
- Therefore, \( \square = 15 \).

21. \( 2^2 + 3^2 = \square \)
- Calculate:
\[
2^2 = 4 \quad \text{and} \quad 3^2 = 9 \quad \implies 4 + 9 = 13
\]
- Therefore, \( \square = 13 \).

22. \( \square^2 \times 2^2 = 100 \)
- Let \( \square = x \). Then:
\[
x^2 \times 4 = 100 \implies x^2 = \frac{100}{4} = 25 \implies x = 5
\]
- Therefore, \( \square = 5 \).

23. \( 2^4 = \square \)
- Calculate:
\[
2^4 = 16
\]
- Therefore, \( \square = 16 \).

24. \( 14 = \square^1 \)
- Any number raised to the power of 1 is itself:
\[
\square = 14
\]

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Final Answers:

\[
\boxed{
\begin{array}{|c|c|c|}
\hline
7 & 11 & 17 \\
\hline
9 & 23 & 1 \\
\hline
12 & 19 & 18 \\
\hline
8 & 21 & 15 \\
\hline
24 & 4 & 13 \\
\hline
10 & 6 & 5 \\
\hline
22 & 2 & 16 \\
\hline
20 & 3 & 14 \\
\hline
\end{array}
}
\]