Binomial theorem proof question.

[Aion]

Definition 1. By the notation $p^k(x)$ we mean the $k$:th derivative of $p$, the polynomial we get by differentiating $p(x)$ $k$ times.

Now suppose $p(x)=x^n$. Then the $k:$th derivative of $x^n$ is

$$\frac{n!}{(n-k)!}x^{n-k}$$
Let $p(x)=a_0+a_1x+...+a_nx^n$ be our usual polynomial. If we differentiate it $k$ times all terms with degree $< k$ vanish and we have

$$p^k(x)=(a_k)k!+a_{k+1}\left(\frac{(k+1)!}{1!}\right)x+...+a_n\left(\frac{n!}{(n-k)!}\right)x^{n-k}$$
If we let $x=0$, then only the constant term "survives",

$$p^k(0)=k!a_k \quad \text{or} \quad a_k=\frac{p^k(0)}{k!}$$Hence we can express the coefficients in $p$ with the help of $p$:s derivative in 0. We can now use this trick on a special polynomial: $p(x)=(1+x)^n$. By a previous theorem, the $k:$th derivative of $p$ is equal to

$$n(n-1)...(n-k+1)(1+x)^{n-k}=\frac{n!}{(n-k)!}(1+x)^{n-k}$$
So
$$p^k(0)=\frac{n!}{(n-k)!}$$
It follows that the coefficient for $x^k$ in $(1+x)^n$ is equal to

$$\frac{p^k(0)}{k!}=\frac{n!}{(n-k)!k!}=\binom{n}{k}$$ Which is the coefficient for $x^k$ in $(1+x)^n$


What confuses me about this proof is that the author uses the same notation for all polynomials. I.e first that $p(x)=(x+1)^n$ and then as $p(x)=a_0+a_1x+...+a_nx^n$. Suppose for clarity that we denote $p_*(x)=a_0+a_1x+...+a_nx^n$. Then the underlying assumption seems to be that

$$p^k(0)=p_*^k(0)$$
This is not entirely obvious to me. Is this true since the polynomial $p_*(x)$ was more general?

 

[Aion]

I think I've figured it out now. Since the relationship $a_k=\frac{p_*^k(0)}{k!}$ holds for any general polynomial $p_*$, it follows that it must be true for the specific polynomial $p(x)=(x+1)^n$ as well.