Let $a_ns^n+\cdot+a_1s+a_0 = 0$ be an equation with roots with negative real part, then $a_i > 0 \forall i$ I have problems in proving the statement

Let $a_ns^n + a_{n-1}s^{n-1} + \cdot + a_1s+a_0 = 0$ be an equation (which coefficients are real) which roots $z_i$ all have a negative real part: $\Re(z_i) < 0 \forall i = 1, \dots, n$, then all the coefficients $a_i$ are positive.
  EDIT: we suppose that $a_n>0$ and $a_0 \ne 0$

What have I tried:
I supposed to have such an equation and to know its roots have negative real parts. So I rewritten the equation in this form: $(s-z_1)(s-z_2)\cdot(s-z_n)$.
Then I said that, if $\mathcal{P}_k$ is the set of all the subsets of the roots of the equation, which cardinality is $k$,  $$a_i = (-1)^i \sum_{\mathcal{Z} \in P_{n-i}} \left( \prod_{z \in \mathcal{Z}} z \right)$$
Now, as $a_i \in \Bbb{R}$, we have $a_i = \Re(a_i)$ so: $$a_i = (-1)^i \Re \left(\sum_{\mathcal{Z} \in P_{n-i}} \left( \prod_{z \in \mathcal{Z}} z \right)\right)$$
Then I got blocked: I do not know how to handle produts of roots (especially non conjugate ones). I tried, without much success, to say that $z_i = \lvert z \rvert e^{i \varphi_z}$ and then $\frac{\pi}{2} \lt \varphi_z \lt \frac{3}{2}\pi$.
Could you please give me some hints for continuoing the proof?
Thanks in advance
 A: (As WE Tutorial School notes, we must assume $a_n>0$ for this to work)
There is a simpler approach.  
If $s+a_0$ is any linear monic polynomial with real coefficients where all roots have a negative real part, then we trivially have that $a_0$ is positive.
Similarly suppose $s^2+a_1s+a_0$ is any quadratic monic polynomial with real coefficients where all roots have a negative real part. Let $x$ and $y$ be the two roots of this polynomial. Given that $s^2+a_1s+a_0=(s-x)(s-y)$, we have $a_1=-x-y$ and $a_0=xy$. If both $x$ and $y$ are real, then they are both negative real numbers so it is immediate that $a_1$ and $a_0$ are positive. If one of $x,y$ is some complex number, $z$, then the other equals $\bar z$, so $a_1=-z-\bar z$ and $a_0=z\bar z$. Again, it is obvious that $a_1$ and $a_0$ are positive. This completes the quadratic case as well. 
Now let $P(s)=a_ns^n+a_{n-1}s^{n-1}+\dots+a_1s+a_0$ be any polynomial with real coefficients, where all roots have a negative real component, and where $a_n>0$.  
We may factor the polynomial as $a_n(s-r_1)(s-r_2)\dots (s-r_k)(s-z_1)(s-\bar z_1)(s-z_2)(s-\bar z_2)\dots (s-z_j)(s-\bar z_j)$ 
where the first $k$ factors are of real roots and the last $j$ factors are of complex roots in conjugate pairs. Note that each product of complex root terms, $(s-z_i)(s-\bar z_i)=s^2-2\mathrm{Re}(z_i)+|z_i|^2$ has real coefficients. So we may rewrite the factorization: 
$a_n(s-r_1)(s-r_2)\dots (s-r_k)(s^2+b_1s+c_1)(s^2+b_2s+c_2)\dots (s-z_j)(s^2+b_ls+c_l)$ 
where each $s^2+b_is+c_i$ has real coefficients (and clearly so does each $s-r_i$). Every root of a term of the form $s-r_i$ or $s^2+b_is+c_i$ is also a root of $P(s)$, so each factor $s-r_i$ or $s^2+b_is+c_i$ has root(s) with negative real part(s).
By what we have already proven, this means each factor ($s-r_i$ or $s^2+b_is+c_i$) has positive coefficients. The product of polynomials with positive coefficients also has positive coefficients. Combined with the fact that $a_n$ is positive, this means $P(s)$ also has all positive coefficients.  
If $a_n$ was negative, then we'd in fact be guaranteed to have a polynomial with all negative roots. This is because then the polynomial $-P(s)$ would have positive leading coefficient $-a_n$, so $-P(s)$ would have all positive coefficients, i.e. $P(s)$ would have all negative coefficients.  
As an example to illustrate this, consider $-3(x+1)(x+2)$. This polynomial satisfies every requirement but that its leading coefficient be positive, and it clearly has all negative coefficients (it equals $-3x^2-9x-6$).
