$n(n+1)(n+2)$ is not a perfect power We have $n,n+1, n+2 \in \mathbb Z^+$ 
Their product can't be a whole exponentiation. Why?

I noticed that $gcd(n,n+1)=1$ and $gcd(n+1,n+2)=1$ This could be a good starting point in the proof. But how do I proceed? Thanks
 A: Because you don't check the links I gave you, I write this answer:
First of all $n+1$ and $n(n+2)=(n+1)^2-1$ are relatively prime, so they are both perfect $k$-th powers for some $k>1$. Let $n+1=a^k$ and $(n+1)^2−1=b^k$. Then $(a^2)^k=1+b^k$. hence 
$$(a^2)^k-b^k=1 $$ but the are not many positive consecutive powers.
Proof by @Andrés Nikolas
A: In 1975, 
Erdos and Selfridge
proved that
the product of consecutive integers
is never a power.
Here is a link to the paper:
http://projecteuclid.org/download/pdf_1/euclid.ijm/1256050816
Here is a link
to the journal where it appeared:
http://projecteuclid.org/euclid.ijm/1256050816
A: $n(n + 1)(n + 2) = n^3 + 3n^2 + 2n$. Obviously this is divisible by $n$. So let's divide it by $n$: we get $n^2 + 3n + 2$. If $n > 2$, then $n^3 - n^2 > 3n + 2$, meaning that $n^2 + 3n + 2$ is between $n^2$ and $n^3$. Therefore $n^2 + 3n + 2$ can't be a perfect power of $n$, and neither can $n^3 + 3n^2 + 2n$.
This still leaves you $n = 1$ and $n = 2$ to check, but that's done easily enough.
