Be smart about this one...

$\prod _{ n=1 }^{ 9 }{ { n }^{ n+2 } } = 1^{3} \times 2^{4} \times 3^{5} \times 4^{6} \times 5^{7} \times 6^{8} \times 7^{9} \times 8^{10} \times 9^{11}$

From the operation, we know that there's prime factors of $2$ and $5$ , resulting $10$ when multiplied, which its units digit is $0$ .

Thus, the units digit of $\prod _{ n=1 }^{ 9 }{ { n }^{ n+2 } }$ is $\boxed {0}$

WOAH...The units, tens, hundredths, thousandths, ten thousandths, hundred thousandths, and the millionths digits are also zero.........by prime factorization.

$\prod_{n=1}^{9} n^{n+2} = 1 ^{3}\times 2^{4} \times 3^{5}\times4^{6}\times 5^{7}\times6^{8}\times7^{9}\times8^{10}\times9^{11}$ By writing out some of the numbers by their factors and adding powers together , we can write this as $2^{54}\times3^{35}\times5^{7}\times7^{9}$ . If you examine the first 8 powers of 2 you can see a pattern. The unit digits always end in 2,4,8 and 6 and this repeats for the rest of the powers of 2. By using sequences you can work out that $2^{54}$ will end with 4 (its unit digit). This can be done for 3,5 and 7 which also have a pattern of repeating unit digits (does not take long to do). The unit digits for $3^{35},5^{7}$ and $7^{9}$ are 7,5 and 7 respectively. Knowing that the unit digits are 4,7,5 and 7 for the above numbers the product is $4\times7\times5\times7 = 980$ which has a unit digit of 0 which is the answer.