Find the number of trailing zeros at the end of the base-$10$ representation of the integer $525^{25^2}
\cdot 252^{52^5}$ .
To find the number of trailing zeros at the end of the base-$10$ representation of the integer $525^{25^2} \cdot 252^{52^5}$, we need to determine the minimum of the powers of $2$ and $5$ in the prime factorization of the number. This is because a trailing zero is produced by a factor of $10$, which is $2 \times 5$.

1. **Prime Factorization of $525$ and $252$:**
   - $525 = 5 \times 105 = 5 \times 5 \times 21 = 5^2 \times 3 \times 7$
   - $252 = 2^2 \times 126 = 2^2 \times 2 \times 63 = 2^2 \times 3^2 \times 7$

2. **Prime Factorization of $525^{25^2}$:**
   - $525^{25^2} = (5^2 \times 3 \times 7)^{25^2} = 5^{2 \cdot 25^2} \times 3^{25^2} \times 7^{25^2}$
   - Therefore, the power of $5$ in $525^{25^2}$ is $2 \cdot 25^2$.

3. **Prime Factorization of $252^{52^5}$:**
   - $252^{52^5} = (2^2 \times 3^2 \times 7)^{52^5} = 2^{2 \cdot 52^5} \times 3^{2 \cdot 52^5} \times 7^{52^5}$
   - Therefore, the power of $2$ in $252^{52^5}$ is $2 \cdot 52^5$.

4. **Combining the Factors:**
   - The total power of $2$ in the product $525^{25^2} \cdot 252^{52^5}$ is $2 \cdot 52^5$.
   - The total power of $5$ in the product $525^{25^2} \cdot 252^{52^5}$ is $2 \cdot 25^2$.

5. **Finding the Minimum Power:**
   - To determine the number of trailing zeros, we take the minimum of the powers of $2$ and $5$.
   - $2 \cdot 52^5$ and $2 \cdot 25^2$ need to be compared.
   - Clearly, $2 \cdot 25^2 = 2 \cdot 625 = 1250$.
   - $2 \cdot 52^5$ is a much larger number than $1250$.

6. **Conclusion:**
   - The number of trailing zeros is determined by the smaller power, which is $2 \cdot 25^2 = 1250$.

The final answer is $\boxed{1250}$