\begin{problem} For $n\in\N$, let $\sigma_k(n)$ be defined as the sum of the $k$th power of the positive divisors of $n$, i.e.: \[ \sigma_k(n) = \sum\limits_{d|n} d^k. \] Let $p$ be prime and $e\in \N_0$. Through experimentation, develop conjectures for the following: \begin{enumerate} \item $\sigma_0(p^e)$ \item $\sigma_1(p)$ \item $\sigma_1(p^e)$ \end{enumerate} Then, choose one of the above to prove.\end{problem}\begin{proof}\end{proof}

This problem is asking for conjectures and a proof for three different mathematical expressions involving the sum of the powers of the divisors of a number.

1. $\sigma_0(p^e)$: The function $\sigma_0(n)$ is the sum of the 0th power of the divisors of $n$. Since any number to the power of 0 is 1, this is equivalent to the number of divisors of $n$. For a prime number $p$ raised to the power $e$, there are $e+1$ divisors (1, $p$, $p^2$, ..., $p^e$). So, we can conjecture that $\sigma_0(p^e) = e+1$.

2. $\sigma_1(p)$: The function $\sigma_1(n)$ is the sum of the 1st power (i.e., the numbers themselves) of the divisors of $n$. For a prime number $p$, the only divisors are 1 and $p$. So, we can conjecture that $\sigma_1(p) = p + 1$.

3. $\sigma_1(p^e)$: This is the sum of the divisors of $p^e$. The divisors of $p^e$ are 1, $p$, $p^2$, ..., $p^e$, so we can conjecture that $\sigma_1(p^e) = 1 + p + p^2 + ... + p^e$.

Let's prove the first conjecture, $\sigma_0(p^e) = e+1$:

Proof: The divisors of $p^e$ are all of the form $p^i$ for $0 \leq i \leq e$. Each $i$ from 0 to $e$ gives a unique divisor, so there are $e+1$ divisors in total. Therefore, $\sigma_0(p^e) = e+1$.