Largest nth Power which has n Digits

Theorem
The largest $n$th power which has $n$ digits is $9^{21}$:


 * $9^{21} = 109 \, 418 \, 989 \, 131 \, 512 \, 359 \, 209$

Proof
The $n$th power of $10$ has $n + 1$ digits.

Hence the $n$th power of $m$ such that $m > 10$ has more than $n$ digits.

The $11$th power of $8$ has $10$ digits:
 * $8^{10} = 8 \, 589 \, 934 \, 592$

and so $8^n$ where $n > 10$ has fewer than $n$ digits.

Hence the $n$th power of $m$ such that $m < 9$ and $n < 21$ has fewer than $n$ digits.

Finally we note that:
 * $9^{22} = 984 \, 770 \, 902 \, 183 \, 611 \, 232 \, 881$

has $21$ digits.

Hence the $n$th power of $9$ such that $n > 21$ has fewer than $n$ digits.

Hence the result.