Proof that the product of 3 sequential numbers is divisible by 3 I am starting to learn about proofs and I ran into this problem:

Prove that the product of 3 sequential numbers is divisible by 3

I am going to present my thoughts on how to prove that and any feedback about whether it is wrong or not would be very appreciated. Thank you in advance
Lets fisrt prove that n³-n is divisible by 3 using Induction:

Constraints
n ∈ Z - {-1, 0, 1}
m ∈ Z

When n = 2,

n³ - n = 6 -> Proved for the base case

n = k
Then k³ - k = 3m

Lets proof for (k + 1)
(k + 1)³ - (k + 1)
=  k³ - k + 3k² + 3k
= 3m + 3k² + 3k
= 3(k² + k + m) -> Proved

Now, lets prove the conjecture of this question by deduction, which is:

Prove that the product of 3 sequential numbers is divisible by 3

(n - 1)(n)(n + 1)
= (n² - n)(n + 1)
= n³ - n
Because It was previously proved that n³ - n is in fact divisible by 3, then it is proved as well
 A: Yeah, it is ok, but you can cut it down drastically. 
Notice that among $3$ consecutive numbers, exactly one is divisible by $3$ (look at their remainder when divided by $3$). So their product is divisible by $3$, and you are done.
A: Since the OP is "starting to learn about proofs" it is worthwhile to expand Aqua's answer with more detail.
Let $m$ be any integer. Using Euclidean_division, we can write
$\tag 1 m = 3q + r \text{ where } r \in \{0,1,2\}$
We want to show that
$\tag 2 n = m (m+1) (m+2)$
is divisible by $3$.
If any one of the three factors on the rhs of $\text{(2)}$ is divisible by $3$, then $n$ is divisible by $3$.
Case 1: r = 0. Then $m = 3q$ is divisible by $3$.
Case 2: r = 1. Then $m + 2 = (3q+1)+2 = 3(q+1)$ is divisible by $3$.
Case 3: r = 2. Then $m + 1 = (3q+2)+1 = 3(q+1)$ is divisible by $3$.
A: Let 3 sequential integer numbers be k-1,k&k+1
Now, k(k-1)(k+1)=k^3-k
We know for any integer k={0,1,2}(mod 3)
Case 1: if, k=0(mod3)
              Then, k^3-k=0(mod 3)
Case 2:if, k=1(mod 3) ,let k=3n+1
             Then, k^3-k=(27n^3+27n^2+9n+1)-(3n+1)=0(mod 3)
Case 3:if, k=2(mod 3) ,let k=3n+2
            Then,k^3-k=(27n^3+54n^2+36n+8)-(3n+2)=0(mod 3)
We see in each case the product is divisible by 3.
So,we're done and hence proved.
