"A polynomial gives rise to the zero function if and only if it is the zero polynomial" In Jim Hefferon's free book on Linear Algebra, there is an exercise that mentions the following:

Prove that a polynomial gives rise to the zero function if and only if it is
  the zero polynomial. (Comment. This question is not a Linear Algebra matter
  but we often use the result. A polynomial gives rise to a function in the natural
  way: $x \mapsto c_n x^n + \cdots + c_1x + x_0$.)

And the proof given is the following:

In this ‘if and only if’ statement, the ‘if’ half is clear—if the polynomial is
  the zero polynomial then the function that arises from the action of the polynomial
  must be the zero function $x \mapsto 0$. For ‘only if’ we write $p(x) = c_n x_n + \cdots + c_1 x + c_0$.
  Plugging in zero $p(0) = 0$ gives that $c_0 = 0$. Taking the derivative and plugging in
  zero $p^{\prime}(0) = 0$ gives that $c_1 = 0$. Similarly we get that each $c_i$ is zero, and $p$ is the zero polynomial.

The ‘if’ half is indeed clear. However, the ‘only if’ part baffles me. How are derivatives related to all of this? Can anyone please shed some light?
 A: This depends heavily on the field over which we're working. For example, over the field with two elements $\;\Bbb F_2\;$ , we have that
$$f(x):= x^2+x\in\Bbb F_2[x]$$
is the zero function, yet it is not the zero polynomial.
The claim is true for any infinite field $\;\Bbb F\;$, though: if $\;f(x)=0\;\;\forall\,x\in\Bbb F\;$ , and if we assume $\;\deg f=n\;$ ,then the polynomial vanishes in more than $\;n\;$ elements, from which it follows it is the zero polynomial. Try to work out carefully the details.
BTW, over any finite field we can always find a non-zero polynomial that, as a function, is the zero function.
A: If $p$ gives rise to the constant function, then all the derivatives of $p$ must vanish at the origin (and everywhere else, in fact). Then taking successive derivatives and evaluating at $0$ shows that all $c_i$ are zero.
A: $$p(x) = c_n x^n + \cdots + c_1 x + c_0$$
$$p'(x) = n c_n x^{n-1} + (n-1)c_{n-1}x^{n-2}\cdots + 2c_2x + c_1 $$
A: Note that the derivative decreases the exponent for each power of $x$.
If
$$p(x)=c_0+c_1x+\cdots+c_nx^n$$
Then
$$p'(x)=c_1+2c_2x+3c_3x^2+\cdots+nc_nx^{n-1}$$
And
$$p''(x)=2c_2+6c_3x+12c_4x^2+\cdots$$
At each point we can conclude that the constant term, hence the coefficient, is 0.
