Find the Taylor series for an arbitrary polynomial

I have to find the Taylor series around $$0$$ for an arbitrary polynomial defined as $$p(x) = a_0 + a_1x + ... + a_{k-1}x^{k-1} + a_kx^k$$ By a few calculations I realized that the Taylor series for the polynomial is just the polynomial it self which makes sense as the idea behind the Taylor series is to find a polynomial which best approximates a given function. But do I have to find this Taylor series by induction? Or is it enough to realize by a few calculations that $$p'(0) = a_1, p''(0) = 2 \cdot a_1 = 2! a_1, ..., p^{(n)}(0) = n! \cdot a_k$$ I am only asked to find the Taylor series and not to use induction but I think I still have to. Can you clarify?

Intuitively, the Taylor series of a polynomial centered at $$0$$ is just the polynomial itself. You verified this by taking derivatives. However, you could also use induction. Given

$$p(x)=\sum_{j=0}^n a_jx^j$$

we need to show that for $$0\le k \le n$$:

$$p^{(k)}(0)=k!a_k$$

where $$p^{(k)}$$ denotes the $$k^{th}$$ derivative with respect to $$x$$. To do this, you can show that the $$k^{th}$$ derivative of $$x^j$$ is

$$\quad\displaystyle\frac{d^{(k)}}{dx^{(k)}}x^j\ =\ \begin{cases}\frac{j!}{(j-k)!}x^{j-k} & \ \text{if}\ k\le j\\ 0 & \ \text{if}\ k>j \end{cases}$$

from which it is clear that $$\quad\displaystyle\frac{d^{(k)}}{dx^{(k)}}x^j\Big|_{x=0}\ =\ \begin{cases}k! & \ \text{if}\ k= j\\ 0 & \ \text{if}\ k\neq j \end{cases}$$

and therefore $$p^{(k)}(0)\ =\ k!a_k$$.

Here's an approach. We know that the Taylor series for $$p(x)$$ around $$0$$ is given by $$\sum_{j=0}^{\infty}\frac{p^{(j)}(0)}{j!}x^j$$. Notice that if $$j>k$$, then we have $$p^{(j)}(x)\equiv 0$$, so this reduces the Taylor series to just $$\sum_{j=0}^{k}\frac{p^{(j)}}{j!}x^j$$. Now for any $$0\leq j\leq k$$, we have, as you pointed out, $$p^{(j)}(0)=j!a_j$$, so we are good to go. Direct calculation works the best here, because with induction you would have to find the higher derivatives at zero assuming the values of the lower derivatives at zero, which is not straightforward.