More odd patterns and even too!

Working modulo 5 (and modulo $\phi(5)=4$ in the exponents), the given number $n$ reduces to $2^2*3^3*2^3*4*3*4^2*2=2^{12}*3^4\equiv{1}\pmod{5}$ . Since $n$ is even, the last digit must be $\boxed{6}$

the last digit will be modeled by the last digit of $2^{n2}\times3^{n3}\times4^{n4}\times6^{n6}\times7^{n7}\times8^{n8}\times9^{n9}$ where $n3$ and $n7$ are divisible by 4, and $n2$ and $n8$ are even but not divisible by 4, and $n4$ and $n9$ are even, thus the last digit of the above product reduces to $4\times1\times6\times6\times1\times4\times1=6.$ The cyclical patterns of the last digit of the powers of each digit are as follows: $2->(2,4,8,6)$ , $3->(3,9,7,1)$ , $4->(4,6)$ , $5->(5)$ , $6->(6)$ , $7->(7,9,3,1)$ , $8->(8,4,2,6)$ , $9->(9,1)$