Show that if $\prod\limits_{k=1}^{n}(x+a_k)=\sum\limits_{k=0}^{n} {n\choose k}a^k_kx^{n-k}$ then $a_1=a_2=a_3=....=a_{n-1}=a_n$ 
Let $a_0=1$. Prove that, if 
  $$\prod_{k=1}^{n}(x+a_k)=\sum_{k=0}^{n} {n\choose k}a^k_kx^{n-k}=x^n+{n\choose 1}a_1x^{n-1}+{n\choose 2}a^2_2x^{n-2}+....+a^n_n,$$
  then $a_1=a_2=a_3=....=a_{n-1}=a_n$.

I proved for $n=2,3$ but I do not know how to prove for general solution. Any help for general solution will be appreciated .

Solution for $n=2$
$a_0=1$
$$2a_1=a_{1}+a_{2}$$
$$a^2_2=a_{1}a_{2}$$
$$a_2=a_{1}$$

Solution for $n=3$
$a_0=1$
$$3a_1=a_1+a_2+a_3$$
$$3a^2_2=a_1a_2+a_1a_3+a_2a_3$$
$$a^3_3=a_1a_2a_3$$
.
.
$$2a_1=a_2+a_3 \tag1$$
$$3a^2_2=a_1a_2+a_1a_3+a_2a_3 \tag2$$
$$a^2_3=a_1a_2 \tag3$$
.
$$3a^2_2=a_1(a_2+a_3)+a_2a_3=2a^2_1+a_2a_3$$
$$6a^2_2=4a^2_1+2a_2a_3=(a_2+a_3)^2+2a_2a_3$$
$$5a^2_2-4a_2a_3-a^2_3=0$$
$$(5a_2+a_3)(a_2-a_3)=0$$
If we select $a_3=-5a_2$ then
$25a_2=a_1$
To check Equation 1
$$2a_1=a_2+a_3$$
$$50a_2 \neq a_2-5a_2$$
Thus $$(5a_2+a_3)(a_2-a_3)=0$$
We must use 
$a_2-a_3=0$
$a_2=a_3$
To check Equation 1,3
Equation 1:
$$2a_1=a_2+a_3$$  
$$a_1=a_3$$
Equation 3:
$a^2_3=a_1a_2 $
$$a_1=a_3$$
Thus 
$$a_1=a_2=a_3$$
 A: If all $a_i$ can be assumed to be positive, then the assertion in question:
$$\prod_{k=1}^{n}(x+a_k)=\sum_{k=0}^{n} {n\choose k}a^k_kx^{n-k}\tag{*1}$$
is a trivial consequence of Maclaurin's inequality (which can be proved using the Newton's inequalities ) :

For any $x_1, x_2, \ldots x_n > 0$ and $k = 1,\ldots, n$, if one
  define the $k^{th}$ symmetric average as:
$$S_k = \frac{\sum_{1\le i_1 < i_2 < \cdots < i_k \le n} x_{i_1} x_{i_2} \cdots x_{i_k}}{\binom{n}{k}}$$
then $S_k$ satisfies following chain of inequalities:
$$S_1 \ge \sqrt{S_2} \ge \sqrt[3]{S_3} \ge \cdots \ge \sqrt[n]{S_n}$$

For the assertion $(*1)$, it is clear $S_k = a_k^k$ and the chain of inequalities become:
$$a_1 \ge a_2 \ge \cdots \ge a_n\quad\implies\quad a_1a_2\cdots a_n \ge a_n^n\tag{*2}$$
It is clear the inequality in $(*2)$ is strict unless all $a_i$ equal to each other. However, the constant term in $(*1)$ tell us $a_1 a_2 \cdots a_n = a_n^n$. From this
we can conclude $a_1 = a_2 = \cdots = a_n$.
I have scratched my head for a few hours and still haven't figured out a way to extend this
argument for general $a_i$. I'm not even sure the $(*1)$ is true when the $a_i$ has mixed signs.
A: 
The triangular inequality, nothing but (the equality case of) the triangular inequality...

Here are some notations. Fix some complex numbers $(a_i)_{1\leqslant i\leqslant n}$ such that the hypothesis holds. Let $A=\max\{|a_i|\,;\,1\leqslant i\leqslant n\}$. Choose some $1\leqslant K\leqslant n$ such that $|a_K|=A$. If $A=0$ then $a_i=0$ for every $i$ hence the result holds. We assume without loss of generality that $A\gt0$. Yet another notation: for every $I\subseteq\{1,2,\ldots,n\}$, let $a(I)=\prod\limits_{i\in I}a_i$, hence $|a(I)|\leqslant A^{K}$ for every $I$ of size $K$.
The hypothesis says that
$$
{n\choose K}a_K^K=\sum_{|I|=K}a(I).
$$
By the triangular inequality,
$$
{n\choose K}A^K={n\choose K}|a_K^K|=\left|\sum_{|I|=K}a(I)\right|\leqslant\sum_{|I|=K}|a(I)|\leqslant{n\choose K}A^K.
$$
Thus, both inequalities above are in fact equalities. 


*

*Since the first inequality is an equality, there exists $z$ such that $|z|=1$ and $a(I)=z|a(I)|$ for every $I$ such that $|I|=K$. 

*Since the second inequality is an equality, $|a(I)|=A^K$ for every $I$ such that $|I|=K$. 


Since $|a_i|\leqslant A$ for every $i$, item 2. yields $|a_i|=A$ for every $i$ (remember that $K\geqslant1$ hence every $i$ belongs to some $I$ such that $|I|=K$). This shows that every $1\leqslant K\leqslant n$ fulfills the hypothesis that $|a_K|=A$, in particular the reasoning above holds for $K=1$. Using item 1. for $K=1$ yields $a_i=zA$ for every $1\leqslant i\leqslant n$, QED.
Edit: This proves the result for $(a_i)_{1\leqslant i\leqslant n}$ in any normed algebra such that the equality case used in the triangular inequality holds, replacing everywhere the modulus $|\ |$ by the norm on the algebra.
A: [From a comment by the OP: "Sorry about the incorrect answer."]
Let $f_1(a_1, a_2, \ldots, a_n; x) = \prod_{i=1}^n (x + a_i)$ and $f_2(a_1, a_2, \ldots, a_n; x) = \sum_{k=0}^n {n \choose k} a_k^k x^{n-k}$. The premise reads that 
$$f_1(a_1, a_2, \ldots, a_n; x) - f_2(a_1, a_2, \ldots, a_n; x) = 0\quad \forall x\quad (*)$$
Denote $\{a_i\}$ as the solution so that (*) is satisfied. Swapping $a_i$ and $a_j$, we have $$f_1(\ldots, a_i, \ldots, a_j, \ldots) = f_1(\ldots, a_j, \ldots, a_i, \ldots)$$ since $f_1$ is symmetric. On the other hand, 
$$f_2(\ldots, a_i, \ldots, a_j, \ldots) - f_2(\ldots, a_j, \ldots, a_i, \ldots) = {n \choose i}(a^i_i - a^i_j)x^{n-i} - {n \choose j}(a^j_i - a^j_j)x^{n-j}\quad (**)$$
From ( * ), (** ) has to be zero for any $x$. This means all of its coefficients have to be zero, which means $a_i^i = a_j^i$ and $a_i^j = a_j^j$. In particular, for $i = 1$ we have $a_1 = a_j, \forall j$ and the proof is complete. 
A: This is an attempted proof by induction.
I am not sure that it is correct.
Let
$f_n
=f_n(a_1, a_2, \ldots, a_n; x) 
=\prod_{i=1}^n (x + a_i)
$
and
$g_n
=g_n(a_1, a_2, \ldots, a_n; x) 
= \sum_{k=0}^n {n \choose k} a_k^k x^{n-k}
= \sum_{k=0}^n {n \choose k} a_{n-k}^{n-k} x^{k}
$.
Then
$f_n' = ((x-a_n)f_{n-1})'
=f_{n-1}+(x-a_n)f_{n-1}'
$
and
$\begin{align}
g_n' 
&= \sum_{k=1}^n {n \choose k} a_{n-k}^{n-k}k x^{k-1}\\
&= \sum_{k=1}^n \left(\dfrac{n!}{k!(n-k)!}\right) a_{n-k}^{n-k}k x^{k-1}\\
&= \sum_{k=1}^n \left(\dfrac{n!}{(k-1)!(n-k)!}\right) a_{n-k}^{n-k} x^{k-1}\\
&= n\sum_{k=0}^{n-1} \left(\dfrac{(n-1)!}{(k)!(n-1-k)!}\right) a_{n-1-k}^{n-1-k} x^{k}\\
&= n\sum_{k=0}^{n-1} {n-1 \choose k} a_{n-1-k}^{n-1-k} x^{k}\\
&= n\sum_{k=0}^{n-1} {n-1 \choose k} a_{k}^{k} x^{n-1-k}\\
&= n g_{n-1}(a_1, ..., a_{n-1}; x)\\
&= n g_{n-1}\\
\end{align}
$
(Start of the induction)
If $f_n=g_n$ 
and
$f_{n-1}=g_{n-1}$
and
$f_n' = g_n'$,
then
$f_{n-1}+(x-a_n)f_{n-1}'
=n g_{n-1}
$
or
$(x-a_n)f_{n-1}'
=(n-1)g_{n-1}
$.
If
$f_{n-1}=g_{n-1}$
implies
$a_i=a_1=a$ for $i=1 ... n-1$,
then
$f_{n-1}' = ((x+a)^{n-1})'
=(n-1)(x+a)^{n-2}
$
and
$g_{n-1}
=\sum_{k=0}^{n-1} {n-1 \choose k} a^{k} x^{n-1-k}
= (x+a)^{n-1}
$
so
$(x+a_n)(n-1)(x+a)^{n-2}
=(n-1)(x+a)^{n-1}
$
so $x+a_n = x+a$
or
$a_n = a$.
