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To solve the problem, let's compute \( \phi(p^k) \) for \( p = 2, 3, 5, 7, 11 \), and \( k = 1, 2, 3, 4, 5 \). The function \( \phi(n) \) (Euler's totient function) counts the number of integers up to \( n \) that are relatively prime to \( n \). For a prime power \( p^k \), the formula is:

\[
\phi(p^k) = p^k \cdot \left(1 - \frac{1}{p}\right) = p^k - p^{k-1}.
\]

### Computation of \( \phi(p^k) \):

#### For \( p = 2 \):
\[
\phi(2^1) = 2^1 - 2^0 = 1
\]
\[
\phi(2^2) = 2^2 - 2^1 = 2
\]
\[
\phi(2^3) = 2^3 - 2^2 = 4
\]
\[
\phi(2^4) = 2^4 - 2^3 = 8
\]
\[
\phi(2^5) = 2^5 - 2^4 = 16
\]

#### For \( p = 3 \):
\[
\phi(3^1) = 3^1 - 3^0 = 2
\]
\[
\phi(3^2) = 3^2 - 3^1 = 6
\]
\[
\phi(3^3) = 3^3 - 3^2 = 18
\]
\[
\phi(3^4) = 3^4 - 3^3 = 54
\]
\[
\phi(3^5) = 3^5 - 3^4 = 162
\]

#### For \( p = 5 \):
\[
\phi(5^1) = 5^1 - 5^0 = 4
\]
\[
\phi(5^2) = 5^2 - 5^1 = 20
\]
\[
\phi(5^3) = 5^3 - 5^2 = 100
\]
\[
\phi(5^4) = 5^4 - 5^3 = 500
\]
\[
\phi(5^5) = 5^5 - 5^4 = 2500
\]

#### For \( p = 7 \):
\[
\phi(7^1) = 7^1 - 7^0 = 6
\]
\[
\phi(7^2) = 7^2 - 7^1 = 42
\]
\[
\phi(7^3) = 7^3 - 7^2 = 294
\]
\[
\phi(7^4) = 7^4 - 7^3 = 2058
\]
\[
\phi(7^5) = 7^5 - 7^4 = 14406
\]

#### For \( p = 11 \):
\[
\phi(11^1) = 11^1 - 11^0 = 10
\]
\[
\phi(11^2) = 11^2 - 11^1 = 110
\]
\[
\phi(11^3) = 11^3 - 11^2 = 1210
\]
\[
\phi(11^4) = 11^4 - 11^3 = 13310
\]
\[
\phi(11^5) = 11^5 - 11^4 = 146410
\]

### General Formula:
From these computations, we observe that:
\[
\phi(p^k) = p^k - p^{k-1}.
\]

This can also be written as:
\[
\phi(p^k) = p^{k-1} \cdot (p - 1).
\]

This formula is valid for any prime \( p \) and any \( k \geq 1 \).

Would you like me to expand on any part of the solution or explore further patterns?  

### Follow-Up Questions:
1. Can you deduce the derivation of the general formula for \( \phi(p^k) \)?
2. How does this formula relate to the properties of the totient function for composite numbers?
3. Can you verify these results using a programming approach for larger values of \( k \)?
4. How does the formula for \( \phi(p^k) \) simplify the calculation for other applications of number theory?
5. Can you generalize this formula for non-prime bases?

**Tip:** Always verify your computations with small cases to ensure accuracy when using general formulas.

Compute φ(2^k) for k = 1,..., 5. Then repeat these 5 computations for φ(3^k), φ(5^k), φ(7^k), and φ(11^k). Can you see a general formula for φ(p^k) where p is a prime number and k is any power of p?

Explore how to compute φ(p^k) for primes p and powers k using Euler's Totient Function. This solution derives the general formula φ(p^k) = p^k - p^(k-1) and demonstrates computations for primes 2, 3, 5, 7, and 11. Suitable for advanced high school or undergraduate students.

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