$\prod_{k=1}^{n} \Big( k^2 + k \Big)$ Consider this product:
$$\beta_{n} = \prod_{k=1}^{n} \Big( k^2 + k \Big)$$
I have tried applying associativity rule:
$$\beta_{n} = \prod_{k=1}^{n} \Big(k^2 + k\Big) = \prod_{k=1}^{n}k^2 + \prod_{k=1}^{n}k = (n!)^2 + n! = n! (n! + 1)$$
For $n = 4$ this produces output $600$. But it should be $2880$.
WolframAlpha gives this as a result: $\beta_{n} = (n!)^2(n+1)$
Hence I assume that associativity rule does not work like it would work with sums.
So how do I go about computing this quite simple product?
 A: As @Winther mentioned in the comments, "a product of a sum is not the sum of a product": You tried to use a formula of the form $\prod_{k=1}^n(a_k+b_k)=\prod_{k=1}^na_k+\prod_{k=1}^n b_k$, but this is not true even for $n=2$:
$$\prod_{k=1}^2(a_k+b_k)=a_1a_2+a_1b_2+b_1a_2+b_1b_2,\quad\text{whereas}\quad \prod_{k=1}^2a_k+\prod_{k=1}^2 b_k=a_1a_2+b_1b_2.$$

In general, for sequences $(a_k)_k$ and $(b_k)_k$, we have
\begin{align*}
\prod_{k=1}^n a_kb_k
&=(a_1b_1)(a_2b_2)\cdots(a_nb_n)\\
&=(a_1a_2\cdots a_n)(b_1b_2\cdots b_n)\\
&=\left(\prod_{k=1}^n a_k\right)\left(\prod_{k=1}^n b_k\right)
\end{align*}
We apply this for $a_k=k$, $b_k=k+1$:
\begin{align*}
\prod_{k=1}^n(k^2+k)&=\prod_{k=1}^nk(k+1)\\
&=\left(\prod_{k=1}^nk\right)\left(\prod_{k=1}^n(k+1)\right)\\
&=n!(n+1)!\\
&=(n!)^2(n+1)
\end{align*}
A: \begin{align*}
&\prod_{k=1}^{n} \bigl(k^2 + k \bigr)\\[4pt]
=\;&\prod_{k=1}^{n} \bigl(k(k+1)\bigr)\\[4pt]
=\;&\left(\prod_{k=1}^{n} k\right)\!\!\left(\prod_{k=1}^{n} (k+1)\right)\\[4pt]
=\;&n!{\,\cdot\,}(n+1)!\\[4pt]
=\;&n!{\,\cdot\,}\bigl((n+1){\,\cdot\,}n!\bigr)\\[4pt]
=\;&(n+1)(n!)^2\\[4pt]
\end{align*}
As regards your attempt . . .

To see that your attempted simplification
$$\prod_{k=1}^{n} \Big(k^2 + k\Big) = \prod_{k=1}^{n}k^2 + \prod_{k=1}^{n}k$$
is not valid, note that for $n=2$, the LHS evaluates to 
$$(1^2+1)(2^2+2)=(2)(6)=12$$
whereas the RHS evaluates to 
$$(1^2)(2^2)+(1)(2)=4+2=6$$
A: You can't distribute the product:
$$\prod_{k=1}^{n}(a_k+b_k)\ne \prod_{k=1}^na_k+\prod_{k=1}^n b_k.$$
However,
$$\prod_{k=1}^n(a_kb_k)=\left(\prod_{k=1}^n a_k\right)\left(\prod_{k=1}^n b_k\right).$$
That being said, we have
$$\begin{align}
\prod_{k=1}^nk(k+1)&=\left(\prod_{k=1}^n k\right)\left(\prod_{k=1}^n (k+1)\right)\\
&=n!\prod_{k=2}^{n+1} k\\
&=n!(n+1)\prod_{k=2}^{n} k\\
&=n!^2(n+1)
\end{align}$$
