Math Help - Taylor's polynomial, proof for existence and uniqueness

1. Taylor's polynomial, proof for existence and uniqueness

Unfortunately I'm not talking about Taylor's theorem, but Taylor's polynomial. The problem stated is : Let $f$ be a function that can be derived at least $n$ times in $a$. Define with precision what is the Taylor's polynomial of order $n$ of $f$ in $a$. Prove that it exists and that it is unique.
My attempt : I define it : The Taylor's polynomial of order $n$ of $f$ in $a$ is the only polynomial $p(x)$ of degree lesser or equal to $n$ that satisfies $p^k(a)=f^k(a)$, $\forall 0\leq k \leq n$, with $k \in \mathbb{N}$ union $\text { { }0 \text {}}$. Now I have to prove that it exists and that it's unique. How can I prove that?
Looking to my notes, the proof is not done, but it says to write the polynomial this way : Let the polynomial be $p_{n,a}(x)=a_0+a_1(x-a)+a_2(x-a)^2+...+a_n(x-a)^n$, with $a_k=\frac{f^{k}(a)}{k!}$ and then after a few algebra with the derivatives of the polynomial, it says $p^k_{n,a}(a)=f^k(a)\Leftrightarrow k!a_k=f^k(a) \Leftrightarrow a_k=\frac{f^k(a)}{k!}$. The funny part is that now it says, $\Leftarrow$) shows the existence while $\Rightarrow$) shows the uniqueness.
I'm very confused. If I understood well, I must show that if the polynomial can be written such as above, then $a_k=\frac{f^{k}(a)}{k!}$ and that if $a_k=\frac{f^{k}(a)}{k!}$, then the polynomial can be written as above. But I'm not sure, so I'm asking you. And in case of an affirmation, would you help me? I thought about induction but realized it was not worth it.

2. Originally Posted by arbolis
Unfortunately I'm not talking about Taylor's theorem, but Taylor's polynomial. The problem stated is : Let $f$ be a function that can be derived at least $n$ times in $a$. Define with precision what is the Taylor's polynomial of order $n$ of $f$ in $a$. Prove that it exists and that it is unique.
If $f$ is differenciable $n$ times on an interval containing $a$ then we define $\sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k$ to be its $n$-th Taylor polynomial at $a$. It of course exists since $f^{(0)}(a),f^{(1)}(a),...,f^{(n)}(a)$ all exist by hypothesis. It is unique because we given a formula how to compute this polynomial.

3. Thanks ThePerfectHacker, you gave me a very fast and nice way to solve the problem.