Grogg’s favorite positive integer is $n\ge2$, and Grogg has a lucky coin that comes up heads with some fixed probability $p$, where $0<p<1$. Once each day, Grogg flips his coin, and if it comes up heads, he does two things:
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[*] He eats a cookie. [/*]
[*] He then flips the coin $n$ more times. If the result of these $n$ flips is $n-1$ heads and $1$ tail (in any order), he eats another cookie. [/*]
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He never eats a cookie except as a result of his coin flips. Find all possible values of $n$ and $p$ such that the expected value of the number of cookies that Grogg eats each day is exactly $1$.
To find all possible values of \( n \) and \( p \) such that the expected value of the number of cookies that Grogg eats each day is exactly 1, we need to analyze the problem step by step.

1. **First Flip Analysis**:
   - If the first flip lands tails (with probability \( 1-p \)), Grogg eats 0 cookies.
   - If the first flip lands heads (with probability \( p \)), Grogg eats 1 cookie and then flips the coin \( n \) more times.

2. **Subsequent \( n \) Flips Analysis**:
   - The probability that the next \( n \) flips result in \( n-1 \) heads and 1 tail (in any order) is calculated as follows:
     - The probability of getting \( n-1 \) heads and 1 tail in any order is given by the binomial coefficient:
       \[
       \binom{n}{1} p^{n-1} (1-p) = n p^{n-1} (1-p)
       \]
     - Here, \( \binom{n}{1} = n \) accounts for the different positions the single tail can occupy among the \( n \) flips.

3. **Expected Number of Cookies**:
   - If the first flip is heads, the expected number of cookies Grogg eats is:
     \[
     1 + n p^{n-1} (1-p)
     \]
   - Therefore, the total expected number of cookies Grogg eats each day is:
     \[
     (1-p) \cdot 0 + p \cdot \left(1 + n p^{n-1} (1-p)\right) = p + n p^n (1-p)
     \]

4. **Setting the Expected Value to 1**:
   - We set the expected value to 1 and solve for \( n \) and \( p \):
     \[
     p + n p^n (1-p) = 1
     \]
   - Simplifying this equation:
     \[
     p + n p^n - n p^{n+1} = 1
     \]
     \[
     n p^{n+1} - n p^n - p + 1 = 0
     \]
     \[
     (n p^n - 1)(p - 1) = 0
     \]
   - Since \( 0 < p < 1 \), the term \( p - 1 \neq 0 \). Therefore, we must have:
     \[
     n p^n - 1 = 0
     \]
     \[
     n p^n = 1
     \]
     \[
     p = \frac{1}{n^{1/n}}
     \]

5. **Conclusion**:
   - The values of \( n \) and \( p \) that satisfy the condition are:
     \[
     (n, p) = \left(n, \frac{1}{n^{1/n}}\right) \quad \text{for} \quad n \ge 2
     \]

The final answer is \( \boxed{ (n, p) = \left(n, \frac{1}{n^{1/n}}\right) \quad \text{for} \quad n \ge 2 } \)