problem: If $\displaystyle{f(x)=x^{(x+1)}(x+2)^{(x+3)}}$, then find the value of $f(0)+f(-1)+f(-2)+f(-3)$.
solution: Since  $0^z=0$ for any $z>0,\ f(0) =f(-2)= 0$. Since $(-1)^0=1$, \begin{align*}
  f(0)+f(-1)+f(-2)+f(-3)&=(-1)^0(1)^2+(-3)^{-2}(-1)^0 \\
  &=1+\frac{1}{(-3)^2} = \boxed{\frac{10}{9}}.
  \end{align*}