The set $\{1,2,3,\ldots,n\}$, where $n \geq 5$, can be divided into two subsets so that the sum of the first is equal to the product of the second A peer of mine showed me earlier today this problem, taken from a 7th grade math contest : 

Let $A=\{1,2,3,\ldots,n\}$; (where $n \geq 5$) prove that $A$ can be divided into two disjoint subsets such that the sum of the elements in the first subset is equal to the product of the elements in the second subset.

This has been puzzling me for 15 minutes already, but I'm sure there's a simple, straight-forward way to do it since it's a 7th grade problem, albeit I can't see it.
Can anyone shed some wisdom here ?

The sum of all the numbers is
$$
\sum_{k=1}^nk=\frac{n(n+1)}2\tag{1}
$$
We will find $a$ and $b$ so that
$$
\overbrace{\left(\sum_{k=1}^nk\right)-a-b-1}^{\text{sum of all but $a$, $b$, and $1$}}=\overbrace{\vphantom{\left(\sum_1^n\right)}1ab}^{\text{product}}\tag{2}
$$
That is, we need to find $a$ and $b$ so that
$$
\begin{align}
\frac{n(n+1)}2
&=\overbrace{ab}^{\text{product}}+\overbrace{a+b+1}^{\text{sum}}\\
&=(a+1)(b+1)\tag{3}
\end{align}
$$
and $(3)$ can be solved using
$$
(a,b)=\left\{\begin{array}{}
\left(\frac{n-1}2,n-1\right)&\text{if $n$ is odd}\\
\left(n,\frac{n-2}2\right)&\text{if $n$ is even}\\
\end{array}\right.\tag{4}
$$
Thus, the product of $1$, $a$, and $b$ is the sum of the rest of the numbers from $1$ to $n$.
If $n\ge5$, then $a,b\ge2$ so that they don't interfere with $1$.

Examples
With $n=5$, $a=\frac{5-1}2=2$ and $b=5-1=4$ so $1\cdot2\cdot4=3+5$
With $n=6$, $a=6$ and $b=\frac{6-2}2=2$ so $1\cdot2\cdot6=3+4+5$
With $n=7$, $a=\frac{7-1}2=3$ and $b=7-1=6$ so $1\cdot3\cdot6=2+4+5+7$
With $n=8$, $a=8$ and $b=\frac{8-2}2=3$ so $1\cdot3\cdot8=2+4+5+6+7$
