Why is ((m^e mod n)^d mod n) = (m^ed mod n) in RSA?

Question

$$RsaPrivate(RsaPublic(m)) = (m^e \bmod n)^d \bmod n = (m^e)^d \bmod n$$

$$RsaPublic(RsaPrivate(m)) = (m^d \bmod n)^e \bmod n = (m^e)^d \bmod n$$

why is this always true?