Sum and product of n natural numbers is equal Sum and product of $n$ natural numbers is equal. Prove that this sum $\ge$ $n+s$ where $s$ is the smallest natural number satisfying $2^s$ $\ge$  $s+n$.
 A: It is not true for $n=1$ and the number being $1$ (also $0$ if you accept that as a natural).  You don't say if the numbers have to be distinct.  If they can be the same, $2,2$ also fails because $s=3.$   Assuming they have to be distinct, if $n \ge 4$ the product is at least $24$ so you will have no problem.  If $n=2$, call the numbers $x,y$.  We have $x+y=xy, x=\frac y{(y-1)}$ which has no solution in the naturals.  If $n=3$ and $1$ is not among them the product will be too large, so let the numbers be $1,x,y$.  We have $1+x+y=xy, x=\frac {y+1}{y-1}$, and the only solution is $1,2,3$.  This again fails because $s=3$ and $s+n=6$ which is the sum.
A: I think the problem needs a little bit of a polishing, more specific, we can have $s \geq 0$, $n \geq 1$ and all $n$ number(s) must be strictly $>0$. Now ...

For $n=1$, we can consider $s=0$ because $2^s=2^0 \geq 1 + 0=n+s$ and every $a_1 >0 \Rightarrow a_1 \geq 1=n+s$

For $n>1$ and 
$$K=a_1+a_2+...+a_n=a_1\cdot a_2 \cdot ... \cdot a_n \tag{1}$$
let's note by 


*

*$S_{\color{red}{1}}=\left\{ a_k \mid a_k = \color{red}{1}, k=1..n\right\}$

*$S_{\color{red}{2}}=\left\{ a_k \mid a_k \geq \color{red}{2}, k=1..n\right\}$ and 


$S_2$ is not empty, otherwise $a_1=a_2=...=a_n=1$ and then $a_1+a_2+...+a_n=n > 1=a_1\cdot a_2 \cdot ... \cdot a_n$ contradicting the equality assumption. $S_1$ may be empty, like in the following case $\color{red}{2+2=2\cdot 2}$. Both $S_1$ and $S_2$ are disjoint, i.e. $S_1 \bigcap S_2 = \varnothing $.
Then, noting $\color{red}{\left|S_2\right|=k}$, from $(1)$ and this inequality (applied to $\log_2$ translates as $x \leq 2^{x-1}, \forall x\geq 2$)
$$K=\sum\limits_{k=1}^{n}a_k = \sum\limits_{a_k \in S_1}a_k + \sum\limits_{a_k \in S_2}a_k=(n-k)+\sum\limits_{a_k \in S_2}a_k =\\ n + \sum\limits_{a_k \in S_2}(a_k - 1) \geq n + \sum\limits_{a_k \in S_2}\log_2{a_k} \tag{2}$$
and (from $(2)$!)
$$K=\prod\limits_{k=1}^{n}a_k=\prod\limits_{a_k \in S_1}a_k \cdot \prod\limits_{a_k \in S_2}a_k=\prod\limits_{a_k \in S_2}a_k=2^{\left(\sum\limits_{a_k \in S_2} \log_2{a_k}\right)}\geq n + \sum\limits_{a_k \in S_2}\log_2{a_k} \tag{3}$$
One value for $s$ is 
$$s=\left \lfloor \sum\limits_{a_k \in S_2}\log_2{a_k} \right \rfloor +1 $$
And $s=k$ is not sufficient, since it doesn't work for $\color{red}{1+2+3=1\cdot 2 \cdot 3}$

A few examples 
$$2+2=2\cdot 2$$
$$1+2+3=1\cdot 2 \cdot 3$$
$$1+1+2+2+2=1^2\cdot 2^3$$
$$1+1+1+1+1+1+1+1+2+2+2+2=1^8 \cdot 2^4$$
