Show that the product of two consecutive natural numbers is never a square.

$$ n^2<n(n+1)<(n+1)^2 $$

That's all :)

There are no integer number between $n$ and $n+1$.

This is fine if $n=0$ is not allowed (and if $n=0$ is allowed the claim becomes wrong. As you ask for alternative proofs, what can you see from this inequality: $$ n^2<n^2+n<n^2+2n+1$$

It looks like this may be the same proof as the OP, but here's how I'd write it:

Let $n$ and $n+1$ be consecutive natural numbers. Note that they must have disjoint prime factorizations. I.e. if $p|n$, then $p \nmid (n+1)$.

For $n(n+1)$ to be a perfect square, then the power on every prime in its decomposition must be even. However, since $n$ and $n+1$ have disjoint prime decompositions, then this is only possible if the power on every prime in the decomposition of $n$ is even, and similarly for $n+1$. This would imply that both $n$ and $n+1$ are perfect squares, which is impossible since no two consecutive naturals can be perfect squares.

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Granted, this proof is a lot more clumsy than Hagen's and Oleg's.