I was going through the fundamental theorem in Number Theory where any non zero integer n can be represented as a product of distinct primes. A related problem with this theorem is to prove that for every such number, there exists a prime $p$ such that $p< \sqrt n$.
I was wondering if there is any mathematical proof that no prime $p$ exists for the number $n$ such that $p> \sqrt n$.
No.  Consider that the square root of $14$ is about $3.74$ but $14$ has $7$ as a prime factor.  Also consider that any prime number such as $2$ is its own (only) prime factor, and any number greater than $1$ is greater than its square root.  The theorem you have stated is incorrect: $25$ has no prime factor less than $5$, and $3$ has no prime factor less than $1.732$; however, it is true that every composite number has a prime factor less than or equal to its square root.