Well, to start with, the claim is not completely true: If the number contains of only the digits 0 and 9 (such as 9099090), then clearly ignoring all of them gives you a wrong result (namely 0 instead of 9). So the actual rule is: You can ignore a 9 if there's at least one non-zero non-ignored digit remaining.
OK, so how to prove that? Well, since the digits are summed up, clearly the order of the digits does not matter. Therefore we can safely assume that all 9's are at the end of the number. Also, it is clear that if ignoring them does not change the digital sum, also appending them does not change the digital sum (and vice versa). Finally, it is obvious that to prove that appending arbitrary many 9's doesn't change the digital root, we only have to proof that appending one 9 does not change the digital root. The extra condition is then nicely reduced to the condition that the number we append the 9 to is not 0.
Now what happens if we append a 9 to the number? Well, we increase the digital sum by 9. Since we then calculate the digital root of that digital sum, what we actually have to prove that the digital sum doesn't change if we add 9 to the number we calculate the digital root of.
OK, so we can now consider the following cases:
- The last digit is a 0 (note that this implies that the number is at least a two-digit, since we excluded that our number is 0). Then this 0 is replaced by a 9, and therefore the next digital sum will again be increased by 9.
- The last digit is not 0, and the second-to-last digit is not 9 (if the number id single-digit, add a leading 0). Since adding 9 is the same as adding 10 and then subtracting 1, the last digit will be reduced by 1, and the second-to-last digit will be increased by 1. Obviously this doesn't change the sum of the digits.
- The second-to-last digit is a 9, and the last digit is not a 0. Then the second-to-last digit will be changed to 0, and the rest of the number will change as if the second-to-last digit had been removed (carry!). Clearly removing the second-last digit will again lead to one of the cases listed here (and as soon as we either run out of digits or have reduced the finial digit to 0, we will reach one of the first two cases). Since for each step in that loop but the last one, we subtract 9 once, and in the last one we either add 9, or leave it unchanged, in effect we have changed the digital sum by a multiple of 9.
So we have established that adding 9 to the number changes the digital sum by a multiple of 9. And of course this means that subtracting 9 also changes the digital sum by a multiple of 9, and by iterating, that changing the number by a multiple of 9 also changes the digital sum by a multiple of 9. Also, it is obvious that the digital sum will never be zero unless the number itself was zero.
But the digital root is obtained by doing the digital sum over and over again, so changing the number by a multiple of 9 will change the digital root also by a multiple of 9.
Now the digital root of a non-zero number is always a digit other than 0. Now changing that by a non-zero multiple of 9 will not give a non-zero digit, and therefore the digital sum cannot be changed at all.
And since, as we established in the beginning, ignoring 9's is equivalent to changing the digital sum by a multiple of 9, it follows that ignoring 9's (with the additional condition preventing that we get 0) doesn't change the digital root of the number either.