Let $k$ be a positive integer. A sequence of integers $a_1, a_2, \cdots$ is called $k$-pop if the following holds: for every $n \in \mathbb{N}$, $a_n$ is equal to the number of distinct elements in the set $\{a_1, \cdots , a_{n+k} \}$. Determine, as a function of $k$, how many $k$-pop sequences there are.

[i]Proposed by Sutanay Bhattacharya[/i]
1. **Understanding the Problem:**
   We need to determine the number of $k$-pop sequences, where a sequence $\{a_n\}$ is called $k$-pop if for every $n \in \mathbb{N}$, $a_n$ is equal to the number of distinct elements in the set $\{a_1, a_2, \ldots, a_{n+k}\}$.

2. **Initial Observations:**
   - For any $n$, $a_n \leq a_{n+1} \leq a_n + 1$. This is because adding one more term to the set $\{a_1, a_2, \ldots, a_{n+k}\}$ either introduces a new distinct element or it does not.
   - If $a_{i+1} = a_i$, then $a_{k+i+1}$ must be one of the elements in $\{a_1, a_2, \ldots, a_{k+i}\}$.
   - If $a_{i+1} = a_i + 1$, then $a_{k+i+1}$ must be a new distinct element not in $\{a_1, a_2, \ldots, a_{k+i}\}$.

3. **Defining Operations:**
   - Let $+$ denote the operation $a_i \rightarrow a_{i+1}$ when $a_{i+1} = a_i + 1$.
   - Let $\square$ denote the operation $a_i \rightarrow a_{i+1}$ when $a_{i+1} = a_i$.

4. **Sequence Construction:**
   - The sequence $\{a_1, a_2, \ldots, a_{k+1}\}$ determines the entire sequence.
   - If we know the order of $+$ and $\square$ operations in the first $k+1$ terms, we can extend this pattern to the entire sequence.

5. **Counting Distinct Elements:**
   - Let $m$ be the number of $+$ operations in the first $k+1$ terms. Then there are $k-m$ $\square$ operations.
   - The number of distinct elements in the sequence $\{a_1, a_2, \ldots, a_{k+1}\}$ is $m+1$.
   - Therefore, $a_1 = m+1$.

6. **Combinatorial Counting:**
   - For each $m$ (where $0 \leq m \leq k$), there are $\binom{k}{m}$ ways to choose the positions of the $+$ operations.
   - Summing over all possible values of $m$, we get the total number of $k$-pop sequences.

\[
f(k) = \sum_{m=0}^k \binom{k}{m} = 2^k
\]

The final answer is $\boxed{2^k}$.