Arrange the digits from 1 to 9 in such an order that the first two digits form a number that is a multiple of 2, the first three digits form a number that is a multiple of 3, and so on.
Example: 123456789 doesn't work, because 12 is divisible by 2 and 123 is divisible by 3, but 1234 isn't divisible by 4.
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I found it after observing that:
the 5th digit has to be 5, as the number must be divisible by 5.
the even digits have to be even, as the numbers must be divisible by 2.
as a consequence of the previous statement, the odd digits have to be odd.
the 3rd and 4th digits must be 12, 16, 32, 36, 72, 76, 92 or 96, as they have to be odd-even, not 5, and divisible by 4.
in order to fulfill all the previous conditions and divisibility by 8, the 6th, 7th and 8th digits must be 216, 296, 416, 432, 472, 496, 632, 672, 816, 832, 872 or 896.
combining the previous two statements and the statement about the 5th digit, the only possibilities from 3rd to 8th digits are: 125496, 125896, 165432, 165472, 165832, 165872, 325416, 325496, 325816, 325896, 365472, 365872, 725416, 725496, 725816, 725896, 765432, 765832, 925416, 925816, 965432, 965472, 965832 or 965872.
in all the possibilities of the previous statement, 2 and 6 are present. As the 2nd digit is the only even digit outside that range (3rd-8th), it must be 4 or 8.
combining informations from the previous statements, and applying the rule for divisibility by 3, the only possibilities for the first 3 digits are: 147, 183, 189, 381, 387, 741, 783, 789, 981 or 987.
(from the initial 9! = 362880 candidates, now we have only 20 possibilities.)
147258963 (fails /7)
147658329 (fails /6)
183254967 (fails /6)
183654729 (fails /7)
189654327 (fails /7)
189654723 (fails /7)
381254967 (fails /6)
381654729 (our answer)
387254169 (fails /6)
387254961 (fails /6)
741258963 (fails /7)
741658329 (fails /6)
783254169 (fails /6)
783254961 (fails /6)
789254163 (fails /6)
789654321 (fails /7)
981654327 (fails /7)
981654723 (fails /7)
987254163 (fails /6)
987654321 (fails /7)