Grouping the first and last terms and two middle terms gives $\left(x^{2}+5 x+4\right)\left(x^{2}+5 x+6\right)+2019$, which can be simplified to $\left(x^{2}+5 x+5\right)^{2}-1+2019$. Noting that squares are nonnegative, and verifying that $x^{2}+5 x+5=0$ for some real $x$, the answer is 2018. 
What is the least possible value of $$ (x+1)(x+2)(x+3)(x+4)+2019 $$ where $x$ is a real number? 
[1] $$2017$$ 
[2] $$2018$$ 
[3] $$2019$$ 
[4] $$2020$$ 
[5] $$2021$$ 