Is the zero polynomial the only polynomial that vanishes at every point of $\mathbb C$? The zero polynomial has the property that every value it takes  on $\mathbb C$ is zero.
Is the converse true, or are there other   polynomials $f$ such that $ f(x)=0$,  for all $x \in \mathbb{C}$? 
 A: Any nonzero polynomial can be written as
$$
a_0+a_1x+\dots+a_nx^n
$$
with $a_n\ne0$. This $n$ is called the degree of the polynomial.
The zero polynomial is usually assigned degree $-\infty$ (or no degree at all). If we multiply two nonzero polynomials, the degree of the product is the sum of the degrees of the factors.
Theorem. If $r$ is a root of the polynomial $f(x)$, then there exists a polynomial $g(x)$ such that $f(x)=(x-r)g(x)$.
Proof. This is based on the division algorithm, that allows us to write $f(x)=(x-r)g(x)+a$, where $a$ is a constant, because $x-r$ has degree $1$. From the hypothesis $f(r)=0$, we get $(r-r)g(r)+a=0$, that is $a=0$. QED
Theorem. If $r_1,r_2,\dots,r_m$ are pairwise distinct roots of the nonzero polynomial $f(x)$ and $f(x)$ has degree $n$, then $m\le n$.
Proof. This is clear if $n=1$. Suppose we have proved the thesis for all polynomials of degree $n-1$. Since $f(x)=(x-r_m)g(x)$ and $r_1,\dots,r_{m-1}$ are also roots of $f(x)$ they must be roots of $g(x)$, as $r_i-r_m\ne0$ for $i=1,2,\dots,m-1$. Since the degree of $g(x)$ is $n-1$, the induction hypothesis implies $m-1\le n-1$, that is, $m\le n$. QED
In particular, a polynomial of degree $n$ has at most $n$ roots.
Corollary If a polynomial has infinitely many roots, then it is the zero polynomial.
Proof. If the polynomial is nonzero, it has a degree and so finitely many roots. QED
A: Assume 
$$f(z) = a_n z^n + a_{n-1} z^{n-1} + \dots a_2 z^2 + a_1 z + a_0, \quad a_n \ne 0.$$
Then it is easy to see that 
$$\lim_{t \in \mathbb{R}, t \to +\infty} \frac{f(t)}{t^n} = a_n \ne 0,$$ 
contradicting the fact that $f(z)$ is identically zero.
A: If $f$ were a least degree nonzero polynomial with $\,f(\Bbb N) = 0,\,$ then $\,f(0)= 0\,$ hence $\,f(x) = x g(x).\,$  Thus $\,0 = g(1)=g(2) = \ldots,$ so $\,\bar f(\Bbb N) = 0\,$ for $\bar f(x) = g(x\!+\!1),\,$ contra minimality of $\deg f$.
Remark $\ $ See also the related thread Apostol proof divides by zero?
A: How about: $${\bf 0}(x) = 0 + 0x + 0x^2 + \cdots 
+0x^n\,?$$


How can we prove that ONLY the zero polynomial works?

Simple. Write $p(x) = a_0 + a_1x + \cdots +a_nx^n$. You know that $p(x) = 0$ for all $x \in \Bbb C$. Take $x = 0$. Then you get $a_0 = 0$. Then take $x = 1,2 \ldots$, and solve the system for $a_1,\ldots, a_n$. The system will have the unique solution $a_i = 0, \ \forall \,i$.
