Find the smallest positive integer $k$ such that $k!$ ends in at least $43$ zeroes.
To find the smallest positive integer \( k \) such that \( k! \) ends in at least 43 zeroes, we need to count the number of trailing zeroes in \( k! \). The number of trailing zeroes in \( k! \) is determined by the number of times 10 is a factor in the numbers from 1 to \( k \). Since 10 is the product of 2 and 5, and there are generally more factors of 2 than 5, we only need to count the number of factors of 5 in \( k! \).

The number of factors of 5 in \( k! \) is given by:
\[
\left\lfloor \frac{k}{5} \right\rfloor + \left\lfloor \frac{k}{5^2} \right\rfloor + \left\lfloor \frac{k}{5^3} \right\rfloor + \cdots
\]

We need this sum to be at least 43. Let's start by testing \( k = 175 \) as an estimate.

1. Calculate the number of factors of 5 in \( 175! \):
\[
\left\lfloor \frac{175}{5} \right\rfloor + \left\lfloor \frac{175}{25} \right\rfloor + \left\lfloor \frac{175}{125} \right\rfloor
\]
\[
= \left\lfloor 35 \right\rfloor + \left\lfloor 7 \right\rfloor + \left\lfloor 1.4 \right\rfloor
\]
\[
= 35 + 7 + 1 = 43
\]

Since \( 175! \) has exactly 43 trailing zeroes, we need to check if there is a smaller \( k \) that also results in at least 43 trailing zeroes.

2. Check \( k = 174 \):
\[
\left\lfloor \frac{174}{5} \right\rfloor + \left\lfloor \frac{174}{25} \right\rfloor + \left\lfloor \frac{174}{125} \right\rfloor
\]
\[
= \left\lfloor 34.8 \right\rfloor + \left\lfloor 6.96 \right\rfloor + \left\lfloor 1.392 \right\rfloor
\]
\[
= 34 + 6 + 1 = 41
\]

Since \( 174! \) has only 41 trailing zeroes, \( k = 174 \) is not sufficient.

3. Check \( k = 175 \) again to confirm:
\[
\left\lfloor \frac{175}{5} \right\rfloor + \left\lfloor \frac{175}{25} \right\rfloor + \left\lfloor \frac{175}{125} \right\rfloor
\]
\[
= 35 + 7 + 1 = 43
\]

Thus, the smallest \( k \) such that \( k! \) ends in at least 43 zeroes is indeed \( k = 175 \).

The final answer is \( \boxed{175} \).