Proof for lemma for existence of Taylor's Polynomial There is lemma that says, for every $a \in \mathbb{R}$, for every polynomial $P_n(x)$ of degree at most $n$ is possible to write in this form:
$$P_n(x) = \sum_{k=0}^{n}c_{k}(x - a)^k$$
where $c_k = \frac{P_n^{(k)}(a)}{k!}$.
There is proof using induction:
$$P_{n+1}(x) = \sum_{k=0}^{n + 1}c_{k}x^{k} = c_{n + 1}(x - a)^{n + 1} - c_{n + 1}\sum_{k=0}^{n}{n + 1 \choose k}x^{k}(-a)^{n + 1 - k} + \sum_{k=0}^{n}c_{k}x^{k}$$
I can understand right side is equal to the left side:
$$c_{n + 1}(x - a)^{n + 1} - c_{n + 1}\sum_{k=0}^{n}{n + 1 \choose k}x^{k}(-a)^{n + 1 - k} + \sum_{k=0}^{n}c_{k}x^{k} = \\ = c_{n + 1}(x - a)^{n + 1} - c_{n + 1}\left(\sum_{k=0}^{n + 1}{n + 1 \choose k}x^{k}(-a)^{n + 1 - k} - {n + 1 \choose n + 1}x^{n + 1}(-a)^{n + 1 - n - 1}\right) + \sum_{k=0}^{n}c_{k}x^{k} = \\ = c_{n + 1}(x - a)^{n + 1} - c_{n + 1}\left((x - a)^{n + 1} - x^{n + 1}\right) + \sum_{k=0}^{n}c_{k}x^{k} = \\ = c_{n + 1}x^{n + 1} + \sum_{k=0}^{n}c_{k}x^{k} = \sum_{k=0}^{n + 1}c_{k}x^{k}$$
But can't understand proof it's also equal to Taylor's Polynomial. The closest I got is:
$$c_{n + 1}(x - a)^{n + 1} - c_{n + 1}\sum_{k=0}^{n}{n + 1 \choose k}x^{k}(-a)^{n + 1 - k} + \sum_{k=0}^{n}c_{k}x^{k} = \\ = c_{n + 1}(x - a)^{n + 1} + \sum_{k=0}^{n}c_{k}x^{k} - c_{n + 1}{n + 1 \choose k}x^{k}(-a)^{n + 1 - k} = \\ = c_{n + 1}(x - a)^{n + 1} + \sum_{k=0}^{n}c_{k}(x - a)^k\left(\frac{x_k}{(x - a)^{k}} - \frac{c_{n + 1}{n + 1 \choose k}x^{k}(-a)^{n + 1 - k}}{c_{k}(x - a)^{k}}\right)$$
Source: https://www2.karlin.mff.cuni.cz/~stanekj/vyukaLS2021/prednaska_MAII_2021_06_02.pdf - page 5
Thanks
 A: I write the proof step by step. I hope that it would be helpful.
If the induction hypothesis would be true for $n$, then for $n + 1$ we can write:
\begin{align*}
P_{n+1}(x) = \sum_{k = 0}^{n + 1} c_kx^k = c_{n+1}\color{blue}{x^{n+1}} + \sum_{k = 0}^{n}c_kx^k
\end{align*}
But $x^{n + 1} = (x - a)^{n+1} - \sum_{k = 0}^n \binom{n + 1}{k}(-a)^{n + 1 - k}x^k$, so we have:
\begin{align*}
P_{n + 1}(x) &= c_{n + 1}\color{blue}{\left((x - a)^{n + 1} - \sum_{k = 0}^n \binom{n + 1}{k} (-a)^{n + 1 - k}x^k\right)} + \sum_{k = 0}^n c_kx^k\\
&= c_{n + 1}(x - a)^{n + 1} - c_{n + 1}\color{magenta}{\sum_{k = 0}^n} \binom{n + 1}{k} (-a)^{n + 1 - k}x^k + \color{magenta}{\sum_{k = 0}^n} c_kx^k\\
&= c_{n + 1}(x - a)^{n + 1} + \color{magenta}{\sum_{k = 0}^n} \left(c_kx^k - \binom{n + 1}{k}(-a)^{n + 1 - k}c_{n + 1}x^k\right)\\
&= c_{n + 1}(x - a)^{n + 1} + \color{red}{\sum_{k = 0}^n \left(c_k - \binom{n + 1}{k}(-a)^{n + 1 - k}c_{n + 1}\right)x^k}\\
&= c_{n + 1}(x - a)^{n + 1} + \color{red}{Q_n(x)} \tag{$1$}
\end{align*}
Because $Q_n(x)$ is a polynomial of degree at most $n$, from the induction hypothesis we can write
\begin{align*}
Q_n(x) = \sum_{k = 0}^n \frac{Q_n^{(k)}(a)}{k!}(x - a)^k \tag{$2$}
\end{align*}
On the other hand, because $Q_n(x) = P_{n + 1}(x) - c_{n + 1}(x - a)^{n + 1}$ from $(1)$, by differentiating we have
\begin{align*}
\frac{Q_n^{(k)}(x)}{k!} &= \frac{P_{n + 1}^{(k)}(x)}{k!} - c_{n + 1}\frac{\left((x - a)^{n + 1}\right)^{(k)}}{k!}\\
&= \frac{P_{n + 1}^{(k)}(x)}{k!} - c_{n + 1}\frac{(n + 1)n\cdots(n + 1 - k)(x - a)^{n - k + 1}}{k!}
\end{align*}
By putting $x = a$ in the above equation, we get
\begin{align*}
\frac{Q_n^{(k)}(a)}{k!} = \frac{P_{n + 1}^{(k)}(a)}{k!} \tag{$3$}
\end{align*}
Also, it's obvious that $c_{n + 1} = \frac{P_{n + 1}^{(n + 1)}(a)}{(n + 1)!}$. By using $(3)$ and $(2)$ in $(1)$ we get the result:
\begin{align*}
P_{n + 1}(x) &= c_{n + 1}(x - a)^{n + 1} + Q_n(x)\\
&= \frac{P_{n + 1}^{(n + 1)}(a)}{(n + 1)!}(x - a)^{n + 1} + \sum_{k = 0}^n \frac{P_{n + 1}^{(k)}(a)}{k!}(x - a)^k\\
&= \sum_{k = 0}^{n + 1} \frac{P_{n + 1}^{(k)}(a)}{k!}(x - a)^k
\end{align*}
A: Just differentiate
$$P_n(x) = \sum_{k=0}^{n}c_{k}(x - a)^k,$$
you will yield $P_n^{(k)}(a) = k!c_k$.
