Proving any product of four consecutive integers is one less than a perfect square

You might note that for any $N$ at all, $$(N-1)(N+1) = N^2-1$$ and so is one less than a perfect square. I presume you have seen this before.

Then you could take $N = M+1$ in the formula above, and get $$M\cdot(M+2) = (M+1)^2 - 1$$ is one less than a perfect square.

And here you have $M = n^2+3n$, don't you?

Here's another approach you might use, a sort of brute-force approach. You said you got as far as $$(n^2 + 3n)(n^2 + 3n + 2).$$ Continuing to multiply, you get: $$ n^4 + 6n^3 + 11n^2 + 6n.$$ The claim is that that this is one less than a perfect square, or equivalently, that $$n^4 + 6n^3 + 11n^2 + 6n + 1$$ is a perfect square. If this last expression is a perfect square, it must be the square of something of the form $n^2 + an+b$ for some $a$ and $b$. But what are $a$ and $b$?

Squaring $n^2 + an+b$, we get $$n^4 + 2an^3 +(a^2+2b)n^2 + (2ab)n + b^2$$

and equating the coefficients of the two polynomials we get $$\begin{align} 2a & = 6 \\ a^2+2b & = 11\\ 2ab & = 6 \\ b^2 & = 1 \end{align}$$

which we can easily solve to obtain $a=3, b=1$. So putting together the various steps we have $$n(n+1)(n+2)(n+3) = (n^2 + 3n + 1)^2 - 1$$ which is what we were looking for.

Observe that the product $p$ of four consecutive integers can be written as $p=(x-\frac{3}{2})(x-\frac{1}{2})(x+\frac{1}{2})(x+\frac{3}{2})$ where $x=n+\frac{1}{2}$ for some integer $n$. Then $p=(x^2-\frac{9}{4})(x^2-\frac{1}{4}) = (x^2-\frac{5}{4}+1)(x^2-\frac{5}{4}-1) = (x^2-\frac{5}{4})^2-1$. It remains to show that $x^2-\frac{5}{4}$ is an integer, which is an easy bit of arithmetic.