Try this simple proof

what's there to prove...
n(n+1)(2n+1) is 6 times the sum of squares of first n integers which is obviously an integer.Hence n(n+1)(2n+1) is always an integer if n is an integer....

n*(n+1) is divisible by 2(because if n+1 is odd then n is even and if n is odd then n+1 is even).

if n or n+1 is divisible by 3 then the product is divisible by 6.

Otherwise n+2 is divisible by 3 .Say n+2=3k.
Therefore 2n+1 = 3n+3 - (n+2) = 3(n+1-k) (Divisible by 3)

Hence the product is divisible by 6.
Furthermore,

n(n+1)(2n+1)6 = ⁿ⁺¹C₂ + 2*ⁿ⁺¹C₃ which is an integer.
Therefore 6 divides n(n+1)(2n+1).

n even: n divisible by 2
n odd: n + 1 divisible by 2
n = 3*k: n divisible by 3
n = 3*k + 1: 2*n + 1 = 6*k + 3: divisible by 3
n = 3*k + 2: n + 1 = 3*k +3: divisible by 3
Thus the expression must always be divisible by 2 and also by 3-
hence it must be divisible by 2*3 ie 6.