A function $f$ from the positive integers to the nonnegative integers is defined recursively by $f(1) = 0$ and $f(n+1) = 2^{f(n)}$ for every positive integer $n$.  What is the smallest $n$ such that $f(n)$ exceeds the number of atoms in the observable universe (approximately $10^{80}$)?

[i]Proposed by Evan Chen[/i]
1. We start with the given recursive function \( f \) defined as:
   \[
   f(1) = 0
   \]
   \[
   f(n+1) = 2^{f(n)}
   \]

2. We need to find the smallest \( n \) such that \( f(n) \) exceeds \( 10^{80} \). First, we estimate \( 10^{80} \) in terms of powers of 2:
   \[
   10^{80} \approx 2^{240}
   \]
   This approximation uses the fact that \( \log_2(10) \approx 3.32193 \), so:
   \[
   10^{80} = (2^{\log_2(10)})^{80} = 2^{80 \log_2(10)} \approx 2^{80 \times 3.32193} \approx 2^{265.7544}
   \]
   For simplicity, we use \( 2^{240} \) as a rough estimate.

3. We now compute the values of \( f(n) \) for small \( n \) to find when it exceeds \( 2^{240} \):
   \[
   f(1) = 0
   \]
   \[
   f(2) = 2^{f(1)} = 2^0 = 1
   \]
   \[
   f(3) = 2^{f(2)} = 2^1 = 2
   \]
   \[
   f(4) = 2^{f(3)} = 2^2 = 4
   \]
   \[
   f(5) = 2^{f(4)} = 2^4 = 16
   \]
   \[
   f(6) = 2^{f(5)} = 2^{16} = 65536
   \]
   \[
   f(7) = 2^{f(6)} = 2^{65536}
   \]

4. We observe that \( f(6) = 65536 \) is much smaller than \( 2^{240} \), but \( f(7) = 2^{65536} \) is vastly larger than \( 2^{240} \).

5. Therefore, the smallest \( n \) such that \( f(n) \) exceeds \( 10^{80} \) is \( n = 7 \).

The final answer is \( \boxed{7} \).