smc

Let's write out the sum. Our sum is equal to $$1\ln 1+2\ln 2+3\ln 3+4\ln 4+5\ln 5+6\ln 6=$$ $$\ln 1^1+\ln 2^2+\ln 3^3+\ln 4^4+\ln 5^5+\ln 6^6=$$ $$\ln (1^1\times 2^2\times 3^3\times 4^4\times 5^5\times 6^6)$$ Raising $e$ to the power of this quantity eliminates the natural logarithm, which leaves us with $$e^p=1^1\times 2^2\times 3^3\times 4^4\times 5^5\times 6^6$$ This product has $2$ powers of $2$ in the $2^2$ factor, $4\ast 2=8$ powers of $2$ in the $4^4$ factor, and $6$ powers of $2$ in the $6^6$ factor. This adds up to $2+8+6=16$ powers of two which divide into our quantity, so our answer is $\boxed{2^{16}}$