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I know that there are proofs that for any $n\geq 1$ there is at least one prime $p$ satisfying $i\cdot n\leq p\leq (i+1)n$ for $i\in\{1,2,3\}$. But are there any simpler proof(s) that there is always at least one prime $p$ satisfying $10^i<p<10^{i+1}$?

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@MatthewConroy Presumably, since that's the $i=1$ case in the OP. –  Alex Becker Jan 5 '13 at 22:28
    
So, clearly it follows from Bertrand's postulate (there's a prime between $n$ and $2n$ for any $n>1$) that there's a prime between $10^{i}$ and $10^{i+1}$ for any $i \ge 0$. Is the question whether the latter, much weaker, statement has a simpler proof than Bertrand's postulate? –  mjqxxxx Jan 6 '13 at 22:19
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