R O O T
  x      U
 ---------
 T R O U T

For the discussion, I am replacing letter O with a different letter, to avoid confusion with digit 0.

   R A A T
  x      U
 ---------
 T R A U T


In the units column, T times U gives final digit T; and we know U can't be 1.  So find the possible combinations of values for (T,U) that satisfy all the conditions:
U not = 1
T, U different
T times U gives last digit T

if U=2, there is no possible value for T
if U=3, then T can only be 5: (5,3)
if U=4, there is no possible value for T
if U=5, there is no possible value for T
if U=6, then T can be any of 2, 4, or 8: (2,6)  (4,6)  (8,6)
if U=7, then T can only be 5: (5,7)
if U=8, there is no possible value for T
if U=9, then T can only be 5: (5,9)

The possible combinations for (T,U) are
(5,3)
(5,7)
(5,9)
(2,6)
(4,6)
(8,6)

Now try each combination and find the one that does not lead to a contradiction.

(T,U) = (2,6)

   R A A 2
  x      6
 ---------
 2 R A 6 2

Here, 2 times 6 in the units column is 12, giving us 2 in the units column of the product and a carry of 1.  But then in the tens column we have 6 times A plus a carry of 1 giving us digit 6 in the product.  But for any digit A, 6 times A is even, so 6 times A plus 1 is odd.

So (T,U) = (2,6) is not a solution.

(T,U) = (5,7)

   R A A 5
  x      7
 ---------
 5 R A 7 5

Here, 5 times 7 in the units column is 35, so we have 5 in the units column of the product, with a carry of 3.
So in the tens column, we have 7 times A plus a carry of 3 giving digit 7 in the product.  The only possible value for A that gives us that is 2, which gives us a 7 in the tens column of the product and a carry of 1.  So

     1 3
   R 2 2 5
  x      7
 ---------
 5 R 2 7 5

But this gives us, in the hundreds column, 7 times 2 plus 1 giving last digit 5; but we know it has to be 2.

So (T,U) = (5,7) is not a solution.