Multiplicative Persistence #2 We had a challenge on Multiplicative Persistence here.
As a recap, to get a multiplicative persistence of a number, do these steps:

Multiply all the digits of a number (in base \$10\$)
Repeat Step 1 until you have a single digit left.
Then count the number of iterations.

More here on Numberphile:

Numberphile &quot;What's special about 277777788888899?&quot;
Numberphile &quot;Multiplicative Persistence (extra footage)&quot;

See the previous challenge.
Rules
The last challenge was all about seeing the intermediate results, but here:

The input should be a positive integer \$n\$.
The output should be the integer with the greatest multiplicative persistence among numbers from \$0\$ to \$n-1\$, choosing the smallest option in case of a tie.
This is code-golf, so the shortest code wins.

Example
[inp]: 10
[out]: 0

[inp]: 100
[out]: 77

[inp]: 50
[out]: 39

[inp]: 31
[out]: 25

Note: Numbers somewhat randomly chosen.
Good Luck!
Husk, 10 bytes
►ȯLU¡oΠd↔ŀ

Try it online!
Explanation
►ȯLU¡oΠd↔ŀ
        ↔ŀ reverse the range 0..n-1
►ȯ         max-by, returning the last maximal element for the following:
    ¡o      create an infinite list using:
      Πd     product of digits.
   U        longest unique prefix
  L         take its length