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Are there arbitary larger numbers of the form $a^b$ with positive integers $a,b>1$ and $a\ne 0\mod 10$ NOT containing all digits ?

Here :

Biggest powers NOT containing all digits.

is either a very similar or the same question, but the question is so old that I decided to restate it.

My current best example :

The number $1955^{39}$ has $129$ digits, but does not contain the digit $1$.

If we assume that the digits are uniformly distributed in $[0,9]$ and considering the growth of the non-trivial powers, can we expect that we have enough chances to get , lets say , a $1000$-digit example ?

In other words, are there enough non-trivial powers with a given decimal-expansion-length, that we can expect that the answer to the question is yes ?

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    How have you obtained those examples? – 2017-02-23
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    brute force using PARI/GP – 2017-02-23
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    Doesn't $a=10^k-1,b=2$ give you infinitely many? You can likely use the same with $b=3$, and $b=4$. Do you want $b$ to be arbitrarily large? – 2017-02-23
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    @ThomasAndrews Correct, I overlooked this family. Do you know further restrictions to avoid such families ? – 2017-02-23
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    Similar stuff with $b=2$ and $a=c10^k\pm 1$ for reasonably simple $c$ (like $c=1,2,3,4,5.) No idea what you'd need to do in general to rule out these sorts of cases. – 2017-02-23

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