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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.

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    $\begingroup$ The "proof verification" tag is for when you have provided a proof and want to know if it is correct. $\endgroup$ – Thomas Andrews Jan 24 '15 at 17:30
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    $\begingroup$ This question is equivalent to prove that there is infinity of prime numbers! $\endgroup$ – user63181 Jan 24 '15 at 17:30
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    $\begingroup$ If not, then all the prime numbers will be less than a fixed integer. So.. $\endgroup$ – Krish Jan 24 '15 at 17:31
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    $\begingroup$ 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. $\endgroup$ – Tahir Imanov Jan 24 '15 at 17:36
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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 $p_{k+1}>p_k$

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

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  • $\begingroup$ 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. $\endgroup$ – Jack D'Aurizio Jan 24 '15 at 17:57
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Use Bertrand's postulate for the number $n$ or any number greater than it.

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