# 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.

• The "proof verification" tag is for when you have provided a proof and want to know if it is correct. – Thomas Andrews Jan 24 '15 at 17:30
• This question is equivalent to prove that there is infinity of prime numbers! – user63181 Jan 24 '15 at 17:30
• If not, then all the prime numbers will be less than a fixed integer. So.. – Krish Jan 24 '15 at 17:31
• Here is thought, prove that there exist $\; k \in \mathbb{N}$ and $\; k \ge n \;$, where $\; 2^k + 1 \;$ or $\; 2^k - 1 \;$ is prime. – Tahir Imanov Jan 24 '15 at 17:36

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 $p_{k+1}>p_k$

Think about the prime factors of $n!+1$.

• I think this answer is just fine, but maybe it is the case to make it less cryptical since it was marked as low-quality. – Jack D'Aurizio Jan 24 '15 at 17:57

Use Bertrand's postulate for the number $n$ or any number greater than it.