#### Answer

$m^4 + 4 m^3 n^2 + 6 m^2 n^4 + 4 m n^6 + n^8$

#### Work Step by Step

$(x+y)^n=\binom{n}{0}x^ny^0+\binom{n}{1}x^{n-1}y^1+...+\binom{n}{a}x^{n-a}y^a+..\binom{n}{n}x^{0}y^n$
Here: $n=4$, $x=m$, $y=n^2$
$(m+n^2)^4=\binom{4}{0}m^4(n^2)^0+\binom{4}{1}m^{4-1}(n^2)^1+\binom{4}{2}m^{4-2}(n^2)^2+\binom{4}{3}m^{4-3}(n^2)^3+\binom{4}{4}m^{4-4}(n^2)^4=$
$m^4 + 4 m^3 n^2 + 6 m^2 n^4 + 4 m n^6 + n^8$