On the equality of the sum of a set and the product of a set. Questions:

  
*
  
*For every $n \in \mathbb{N}$, does there exist a set $S = \{s_1, s_2, \ldots, s_n\} \subset \mathbb{Q}$, such that $$\prod_{k=1}^{n}s_k = \sum_{k=1}^{n}s_k$$ (EDIT: and the sum and product are non-zero)
  
*If not, can the upper bound for $n$ be shown, such the above is true for $n$ less than the bound? (for the above and/or the following questions)
  
*The same as question one, but with the added restriction that the elements of $S$ are integers?
  
*The same as question one, but with the added restriction that the least element in $S$ is $1$?
  

I came up with this problem when looking at the set $\{1, 2, 3\}$.  I noted that $1+2+3 = 6 = 1\cdot2\cdot3$.  This was quite interesting, and I looked for other sets with this property, and easily noted $\{-1, -2, -3\}$ and the trivial $\{1\}$ and $\{0\}$.  Not so easily noted was $\{\frac{-4}{7}, \frac{-1}{2}, \frac{3}{2}\}$.
Thinking of sets of four or more numbers was more difficult.  But, I noted that $\{\frac{-6}{5}, \frac{-1}{2}, \frac{3}{2}, 2\}$ also satisfies this property.
I was curious if anyone here might have some insight.
EDIT: 
I just realized that the proof of question 1 is simple:
For odd $n$, let the set be $\{0, s_1, -s_1, s_2, -s_2, (etc.)\}$.  Thus, the sums cancel out, and the multiplication cancels because of $0$.  
For even $n$, let the set be similar: $\{0, s_1, s_2, -(s_2+s_1), s_3, -s_3, s_4, -s_4 (etc.)\}$
Thus, let question one have the restriction that the resulting equality cannot be $0=0$, to make it more interesting.
 A: Given (almost) any set of numbers, we can add one more number to it to make a set like you want.  Let $S=\sum s_k, P=\prod s_k$.  Then we can add in $t$ to make $S+t=Pt$ by choosing $t=\frac S{P-1}$ unless $P=1$.  If all the numbers in $S$ are rational, so will $t$ be.  Aside from the product being $1$, this can fail if the desired $t$ is already in the set.
For integers, $\{-n,0,n\}$ qualifies for any $n$ as does $\{-m,-n,0,m,n\}$.  This approach allows arbitrarily large sets.
For sets with minimum $1$ and integers, $\{1,2,3\}$ is the best you will do.  The product gets too large.
A: For 3: If the numbers are distinct positive integers, the $n \leq 3$. Indeed assume by contradiction that $n \geq 3$. Without loss of generality 
$$s_1<s_2<...<s_n \,.$$
This means that $s_2 \geq 2, s_{n-1} \geq n-1$ and hence, 
$$ns_n > s_1+..+s_n =s_1..s_n > s_2s_{n-1}s_n \geq 2(n-1)s_n$$
This shows that $n> 2(n-1)$ contradiction.
For $n=3$ I think the only positive solution is $1,2,3$. [ If $a<b<c$ and $abc=a+b+c<3c$ then $ab <3$. This shows that $a=1, b=2$.]
For $n=2$ the only solutions to $ab=a+b$ are $(2,2), (0,0)$ (you can factor it as $(a-1)(b-1)=1$). But these are not distinct.
If the numbers can be equal, note that if there are exactly n terms , $(1,1,1,..,1,2,n)$ has the same sum and product.... 

For general integers the following Lemma, which can be proven exactly as before, seems to be the key:
Let $b_1,..,b_n$ be distinct non-zero integers of the same sign.


*

*If $n \geq 4$ then $|b_1..b_n| > |b_1|+...|b_n|$.

*If $n =3$ then $|b_1b_2b_3| \geq |b_1|+|b_2|+|b_3|$ with equality only for $\pm1,\pm2,\pm3$.

*If $n =2$ then $|b_1b_2| \leq |b_1|+|b_2|$ implies $|b_1|=1$.


Now splitting the sum into positive and negatives, you seem immediately that you must have at most $3$ positives OR $3$ negatives. In each of $3$ positives, $3$ negatives, $2$ positives, $2$ negatives case you get immediately simple upper bounds for the other sign, and very few solutions.
The case with one negative is easy to treat, so all you have to do is the one positive case. 
I think this can solve the problem, but the number of cases is too high..
