Any natural number is greater than or equal to its product of digits. I was randomly thinking and I stumbled upon this question in my mind. I was thinking of numbers, and I observed that any number is greater than or equal to the product of digits in many cases. I was thinking, is this true for all natural numbers or for only a few cases? Please help.
Let $k$ be the number of digits, and let $a$ be the first digit (assume $a&gt;0$).  Then the product of the digits is bounded above by $a\times 9^{k-1}$.  The value of the number itself is bounded below by the total value of that first digit: $a\times 10^{k-1}$.
Let k be the number of digits, and ak-1 ak-2 .... a0 are the digits of the number.

for example, Let the number is 265 then k=3 , a2 = 2, a1=6, a0 = 5.

Then product of digit of the number is  &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  ak-1 * ak-2 * ... * a0

The actual value of number is &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  ak-1 * 10k-1 + ak-2 * 10k-2 +....+a0 * 100

Compare product of digit of the number and 1st part of actual value of the number

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  ak-1 * ak-2 * ... * a0

&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  ak-1 * 10k-1

In product of digit of the number after first digit (ak-1) there are product of k-1 digit such that 0&lt;=digit&lt;10.

In actual value of number after first digit (ak-1) there are product of k-1  10.

10k-1 always grater than product of k-1 number in {0-9}.

Therefore,

If k>1 the number  is always grater than product of digit of the number.

And if k=1 we can't find the product of digit of the number because product is binary operator.

Any natural number is greater than or equal to its product of digits.