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Today I found an interesting problem on the Internet - to find the exact power that has the largest possible product of digits.

Of course, we know that there are arbitrarily large squares that do not contain zeros, and therefore, for which the product of digits is arbitrarily large.

But this problem would be interesting for an exponent greater than 2.

The question is, is there any way to find arbitrarily large cubes (or other perfect powers) of natural numbers that do not contain a zero in the decimal notation?

The record that I discovered is $58403809^{22}$, that contains 171 decimal digits, and the product of which contains 115 digits.

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According to https://oeis.org/A052427,

"Pegg (1999) conjectured that the sequence of zeroless cubes (A052045) is finite. On April 19, 1999, Hickerson gave the counterexample: if $n \equiv2 \pmod 3$ and $n \ge 5$, then the cube of $(2\times10^{5n} - 10^{4n} + 17\times10^{3n-1} + 10^{2n} + 10^n - 2)/3$ is zeroless. Three days later, Baxter gave a simpler variation which is valid for all $n\ge0$ and is given in the Formula section."

That formula is $$ a(n) = (2\times10^{5n} - 10^{4n} + 2\times10^{3n} + 10^{2n} + 10^n + 1)/3 $$

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