I proved that $$ \{ 0^n 1^{5n} \mid n \geq 0 \}$$  is not a regular language using Pumping Lemma by following way.Solve by contradiction that $ L = \{0^n 1^{5n} \mid n \geq 0 \}$ is regular language.Let $0^p 1^{5p}$ be in L, where $p$ is the pumping length.Now here if the language $L$ is regular language, $0^p 1^{5p}$ can be represented in the form $xyz$ where $|xy| \le p$ & $|y|\gt0$.Thus, from step 2, we can say that $xy = 0^p$ and $y = 0^j$.So, $xyz = (0^{p-j})(0^j)(1^{5p})$.Now pumping the value of y to 2, $xyyz = (0^{p-j})(0^j)(0^j)(1^{5p}) = (0^{p+j})(1^{5p})$, which is surely not in $L$, thus not a regular language.But how to prove for condition $\{0^n 1^{5n} \mid n \ge 10000 \}$ & for also $n \le 10000$, we can just prove that one of them is not regular and obviously by rules the complement will also be not regular.
How to prove that $\{0^n 1^{5n} \mid n \ge 10000 \}$ is not a regular language?
formal languages;regular languages;pumping lemma
The condition $n>10\,000$ makes no real difference. If the pumping length is greater than $10\,000$, your existing proof works. If the pumping length is $10\,000$ or less, then pumping the string $0^{10\,000}1^{50\,000}$ gives you a string that isn't in the language because it has too many zeroes.