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Note: The apostrophes are meant to separate different groups of digits. Like, $0.{1^2}'{2^2}'{3^2}'{4^2}'\cdots=0.14916\cdots$. I wasn't able to come up with something better.

It is easy to show $0.{1^n}'{2^n}'{3^n}'{4^n}'\cdots$ is always irrational: if it were not, we would find that the number of consecutive zeroes in its period would have a maximum, but obviously we can always get one more reaching a new power of $10$. When it comes to the powers of a single natural number $m\ne 0,1$, in general that is $0.{m^0}'{m^1}'{m^2}'\cdots$, the irrationality follows from the fact that the last $1,2,3,4,\cdots,l$ digits of such powers repeat with different periods, yielding no total repetition.

We then have Copeland-Erdos's constant $0.2357\cdots$, whose irrationality follows from Dirichlet's theorem. Does this work for the number $x$ whose decimal expansion is formed by concatenating the $n$-th powers of the primes? Or, do we have a different way?

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    $\begingroup$ What does $0.2^{n'}3 \cdots $ mean? Something like $0.2xxx3$? $\endgroup$ – johannesvalks Aug 24 '15 at 17:36
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    $\begingroup$ @johannes: The apostrophes are meant to separate different groups of digits. Like, $0.{1^2}'{2^2}'{3^2}'{4^2}'\cdots=0.14916\cdots$. I wasn't able to come up with something better. :/ $\endgroup$ – Vincenzo Oliva Aug 24 '15 at 17:40
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    $\begingroup$ Ok thanks - that is clear... Nice question +1!!! $\endgroup$ – johannesvalks Aug 24 '15 at 17:43
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    $\begingroup$ I think that the explanation of the notation should go at the top, not the bottom. $\endgroup$ – Akiva Weinberger Aug 24 '15 at 18:04
  • $\begingroup$ @columbus: Yeah, I agree, thanks. $\endgroup$ – Vincenzo Oliva Aug 24 '15 at 18:05
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By Dirichlet's theorem on primes in arithmetic progressions, for any $k$ there are an infinite number of primes of the form $10^km+1$.

If you choose $k$ large enough compared to $n$, there will be an arbitrarily large number of consecutive zeros. Therefore, the number can not repeat, and is therefore irrational.

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You can have arbitrary large sequence of $0$'s:

Suppose not, there is a period $k$ in the decimal expansion.

By Dirichlet's theorem, we have infinitely many prime numbers $p$ in the form $$ p= a 10^{k+2} + 1.$$ Then

$p^n \equiv 1$ (mod $10^{k+2}$).

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