Answer
Let $n$, $n+1$, $n+2$, and $n+3$ be four arbitrary consecutive integers. Then $n(n+1)(n+2)(n+3)=n(n+1)(n^{2}+5n+6)=n(n^{3}+6n^{2}+11n+6)=n^{4}+6n^{3}+11n^{2}+6n+1-1=(n^{2}+3n+1)^{2}-1$. Since $n^{2}+3n+1$ is an integer (because it is the result of adding and multiplying integers), it must be that $(n^{2}+3n+1)^{2}$ is a perfect square, so $(n^{2}+3n+1)^{2}-1$ is one less than a perfect square. Therefore, the product of any four consecutive integers is one less than a perfect square.
Work Step by Step
Only two steps are really worth paying close attention to. First, we discreetly add $1-1$ (which equals zero) into the equation above because we are trying to write the expression as something squared minus one, but before that, there were no constant terms in the expression. Second, we factor $n^{4}+6n^{3}+11n^{2}+6n+1$ in a single step because there is no work to show: the only way to verify that the factorization is valid is by carefully multiplying out the terms of $(n^{2}+3n+1)^{2}$, or by using a computer algebra system to do so.
