jmc

We see that the given expression is part of the factorization of $x^5-y^5$. Since $x-y\ne 0$, we can write $$x^4+x^{3}y+x^2{y}^2+x{y}^3+y^4=\frac{(x-y)(x^4+x^{3}y+x^2{y}^2+x{y}^3+y^4)}{x-y}=\frac{x^5-y^5}{x-y}.$$Plugging in $x=5$ and $y=4$, we get $$\frac{5^5-4^5}{5-4}=3125-1024=\boxed{2101}.$$