Let $p$ be the [i]potentioral[/i] function, from positive integers to positive integers, defined by $p(1) = 1$ and $p(n + 1) = p(n)$, if $n + 1$ is not a perfect power and $p(n + 1) = (n + 1) \cdot p(n)$, otherwise. Is there a positive integer $N$ such that, for all $n > N,$ $p(n) > 2^n$?
1. **Understanding the function $p(n)$:**
   - The function $p(n)$ is defined recursively with $p(1) = 1$.
   - For $n+1$, if $n+1$ is not a perfect power, then $p(n+1) = p(n)$.
   - If $n+1$ is a perfect power, then $p(n+1) = (n+1) \cdot p(n)$.

2. **Analyzing the growth of $p(n)$:**
   - We need to determine if there exists a positive integer $N$ such that for all $n > N$, $p(n) > 2^n$.
   - To do this, we need to estimate the growth of $p(n)$.

3. **Bounding $p(n)$:**
   - Consider the product of $p(n)$ over all perfect powers up to $n$.
   - Let’s denote the set of perfect powers up to $n$ as $\{a^k \mid a^k \leq n, a \in \mathbb{N}, k \geq 2\}$.
   - The number of perfect powers up to $n$ is approximately $\sqrt{n}$.

4. **Estimating the product of perfect powers:**
   - We can write:
     \[
     p(n) \leq \prod_{a=2}^{\sqrt{n}} (a^2 \cdot a^3 \cdot \dots \cdot a^{\log_a n})
     \]
   - Simplifying the product:
     \[
     \prod_{a=2}^{\sqrt{n}} a^{(\log_a n)^2}
     \]
   - Using the change of base formula for logarithms, $\log_a n = \frac{\log n}{\log a}$, we get:
     \[
     \prod_{a=2}^{\sqrt{n}} a^{\left(\frac{\log n}{\log a}\right)^2}
     \]

5. **Converting to exponential form:**
   - We can express the product in terms of the exponential function:
     \[
     \prod_{a=2}^{\sqrt{n}} e^{\frac{1}{\log a} \log^2 n}
     \]
   - This simplifies to:
     \[
     e^{\sum_{a=2}^{\sqrt{n}} \frac{\log^2 n}{\log a}}
     \]

6. **Estimating the sum:**
   - The sum $\sum_{a=2}^{\sqrt{n}} \frac{1}{\log a}$ is approximately $\sqrt{n} \cdot \frac{1}{\log 2}$, since $\log a$ grows slowly.
   - Therefore:
     \[
     e^{\sqrt{n} \cdot \frac{\log^2 n}{\log 2}}
     \]

7. **Comparing with $2^n$:**
   - We need to compare $e^{\sqrt{n} \cdot \frac{\log^2 n}{\log 2}}$ with $2^n$.
   - Note that $2^n = e^{n \log 2}$.
   - Clearly, $e^{\sqrt{n} \cdot \frac{\log^2 n}{\log 2}} = e^{o(n)}$, which grows much slower than $e^{n \log 2}$.

8. **Conclusion:**
   - Since $e^{o(n)}$ grows slower than $2^n$, there is no positive integer $N$ such that for all $n > N$, $p(n) > 2^n$.

\[
\boxed{\text{No, there is no such } N.}
\]