Clever Yet Simple Polynomial Subsitution

Call $g(x)=f(x)-1$ . This means that the roots of $g(x)$ are $1, 2, 3...2009$ . Since $f(x)$ is monic with degree $2009$ , $g(x)$ is also monic with degree $2009$ , since we are not changing the leading coefficient or degree when subtracting $1$ . Thus, we can factor $g(x)$ as $g(x)=(x-1)(x-2)(x-3)....(x-2009)$ . Thus, $g(2010)=2009!$ , and so $f(2010)=2009!+1$ . Substituting this into the answer format, we find the final answer as $\boxed{2010}$ .

Yes, I realized that $f(x)=(x-1)(x-2)(x-3) \dots (x-2009)+1$ . Thus, $f(2010)$ will just be $2009!+1$ , and we're simply asked to find $2010(1)=\boxed{2010}$ . Great problem @Aayush Gupta ! :D