Prove that for any natural number $n$ there exists a prime number $p$ greater than $n$ Prove that for any natural number n there exists a natural prime number p , such that $ p>n $. How can I prove that ? Thank you.
 A: First thing prove that there are infinite number of primes,you can use the Euclid proof imagine that there are $n$ primes and name the $k$-th prime $p_k$ than the number $t=p_1p_2\cdots p_{n-1}p_n+1$ isn't divisible by any prime $p_k$ where $1\leq k\leq n$ so there is another prime which divides $t$.Now it's easy to show that $p_k> k$ for every $k\in\mathbb{N}$ since the first prime is $p_1=2$ and $p_{k+1}>p_k$
A: Think about the prime factors of $n!+1$.
A: Use Bertrand's postulate for the number $n$ or any number greater than it.
A: We know that there are infinitely many natural prime numbers.
Now, for some natural number n, let there do not exist any natural prime p such that $p \gt n$. This implies that all prime numbers are less than or equal to the natural number n. Since, even the total number of natural numbers less than or equal to n is finite, the number of natural prime numbers must be finite which contradicts the above fact.
Hence our assumption is false, therefore its negation must be true i.e., for every natural number n, there exists atleast one natural prime p such that $p \gt n.$
