A is a root of order of polynomial p iff p(a)=p'(a)= =[p^(k-1)](a)=0

[xsw001]

a is a root of order k of the polynomial p provided that k is a natural number such that p(x)=[(x-a)^k]r(x), r is a polynomial and r(a) not equal to 0.

Prove a is a root of order k of the polynomial p iff p(a)=P'(a)=...=[p^(k-1)](a)=0 and [p^(k)](a) not equal to 0.

Note:
[p^(k-1)](a) := differentiate p(x) k-1 times
[[p^(k)](a) := differentiate p(x) k times

Taylor Polynomial expansion:
Pn(x) = f(a)+f'(a)(x-a)+[f''(a)/2!](x-a)^2+...+[f^n(a)/n!](x-a)^n

Proof:
My first instinct is to let finite series of r(x) to be e^x
Then r(x)=e^x=1+x+x^2/2!+x^3/3!+...+x^n/n!
Let a=0, then p(x)=[(x-a)^k]r(x)= (x^k)(e^x)
Differentiate p(x) k times to conclude p(0)=p’(0)=p’’(0)=…=[p^(k-1)](0)=0, but (p^k)(0)=k! not equal to 0 (I have skipped the steps in this post since it takes too long to type it up)
But how should I prove the other way around?
Suppose p(a)=p'(a)=...=[p^(k-1)](a)=0 and [p^(k)](a) not equal to 0 and prove that a is a root of order of polynomial p? Any suggestions?

 

[lanedance]

how about starting from below & differentiating k times
$$p(x)=(x-a)^k r(x)$$

 

[xsw001]

Yeah, that's what I was thinking originally instead of use a particular example, just generalize it by differentiating k times. I got pretty close the same answer,
p(a)=p’(a)=p’’(a)=…=[p^(k-1)](a)=0, but (p^k)(a)=k!r(a) not equal to 0 since r(a) not equal to 0 and k is natural number.

Should I state in order to prove the reverse order is just to integrating k times to get p(x)=[(x-a)^k]r(x)? So a is also a root of order k of the polynomial p. That would complete the if and only if prove?