Consecutive divisibility of higher powers

Everybody knows that its easy to find consecutive natural numbers which are composite. Now if someone asks you can you find consecutive natural numbers which are divisible by higher order powers the answer is again yes. For example consider number 8 and 9. 4 divides 8 and 3 divides 9. So here he was two consecutive numbers which are divisible by square numbers. Question is : Can you find 3 consecutive numbers which are all divisible by square numbers and the answer is yes. To answer this we create a constructive induction proof
first
let a1,a2,...an be the consecutive natural numbers
let s1,s2,...,sn be the square numbers such that s1|a1, s2|a2  ... sn|an
let L=s1*s2*s3*...*sn
Now we have to come up with sequence of n+1 numbers which are divisible
by n+1 square numbers
let these numbers are
a1+A, a2+A,...,an+A, (an+1)+A
where A=an+1(L+2)L
Now notice that each of these numbers a1+A = a1+(an+1)(L+2)L is divisible by s1
similarly a2+A = a2+(an+1)(L+2)L is divisible by s2
...
a2+A = a2+(an+1)(L+2)L is divisible by sn, because L is a product of s1*s2*..sn and sn|an.
The next step is: Is (an+l)+A is also divisible by some square number ? Because we don't have any idea about s(n+1).
(an+1)+A =(an+1)+(an+1)(L+2)L =(an+1)(1+L^2+2L) =(an+1)(L+1)^2.
Thus a(n+1)+A is divisible by a square number which is (L+1)^2

To extend our example we chose in the beginning that 8 and 9 are consecutive numbers which are divisible by square number. Can we get three consecutive numbers which are divisible by square numbers. So we extend the construction as
L = s1*s2 = 4*9=36
A=(an+1)(L+2)L =10(36+2)36=  13680
a1+A = 8+13680 = 13688 means 4|13688
a2+A =9+13680=13689 means 9||13689
a3+A=10+13680=13690 means (L+1)^2=37^2= 1369|13690

To prove for cubic or higher values the h1,h2,..hn
and terms become like a1+A where A =(an+1)((L+1)^m-1)