Proof that $\overline{P(z)} = P(\overline{z})$ for polynomial $P$ with real coefficients 
Let $$ a_0, a_1, a_2, a_3, \ldots , a_n \quad (n \ge 1)$$ denote real numbers, and let $z$ be any complex number.  With the aid of $$ \overline {z_1 +z_2+ \ldots +z_n} = \overline z_1 +\overline z_2+ \ldots + \overline z_n $$ and $$ \overline {z_1 z_2 \ldots z_n} = \overline z_1 \overline z_2 \ldots  \overline z_n, $$ Show that $$ \overline {a_0 + a_1z+ a_2z^2+ \ldots + a_nz^n} = a_0 +  a_1\overline z + a_2\overline z^2+ \ldots + a_n \overline z^n.$$

I was also told to use $z$ is a real number if and only if $\overline z = z$ and is $z$ is either real or pure imaginary if and only if $ \overline z ^2 = z^2$
Any hints as to where to go?  Does this proof take mathematical induction?
I was thinking I could factor out $a_0, a_1, etc)$ and then use the fact that $z$ is real if and only if $\overline z = z$  Then say by the other statement above about the sum of the conjugates of complex numbers the two must be equivalent.  I feel like this is right but it is not a very elegant proof. 
 A: Isn't it much simpler? You have $$\overline{\sum_{i=0}^n a_iz^i}$$
Which, by the first identity, evaluates to   $$\sum_{i=0}^n \overline{a_iz^i}$$
Now using that if $a_i \in \mathbb{R}$, then $\overline{a_i} = a_i$:
 $$\sum_{i=0}^n a_i\overline{z^i}$$
Now we must show that $\overline{z^i} = \overline{z}^i$. That is true by the second identity. Since $$\overline{z^i} = \underbrace{\overline{z\cdot z \cdot ... \cdot z}}_{i-times}=  \underbrace{\overline{z}\cdot \overline{z} \cdot ... \cdot \overline{z}}_{i-times} = \overline{z}^i$$ we must have 
$$\sum_{i=0}^n a_i\overline{z^i} = \sum_{i=0}^n a_i\overline{z}^i = a_0 + a_1\overline{z}+a_2\overline{z}^2+..+a_n\overline{z}^n$$
And the proof is finished.
A: Here is what you need to show:
$\forall c\in \mathbb R,z\in \mathbb C, c\overline z = \overline {cz}$
$\forall z_1,z_2 \in \mathbb C, \overline z_1 + \overline z_2 = \overline {z_1+z_2}$
$\forall z\in \mathbb C, \overline {z}^n = \overline {z^n}$ 
i.e.
$z = r e^{it}\\
\overline z = re^{-it}\\
\overline {z}^n = r^ne^{-int} = \overline {z^n}\\$
And with all of the above: $p(\overline z) = \overline {p(z)}$
