Prove that the Maclaurin series for any polynomial is the polynomial itself

The Maclaurin series for (1-x)^-1 for |x|<1 is a geometric series .

1+x+x^2+x^3+...........

so the Toylor series for x^-1 at a=1 is

1-(x-1)+(x-1)^2-(x-1)^3+..........

By integrating the above Maclaurin series we find the series for log (1-x)

-x-1/2x^2-1/3x^3-1/4x^4

the series for log x at a = 1 is

(x-1)-1/2(x-1)^2+1/3(x-1)^3-1/4(x-1)^4

for the exponential function e^x at a=0 is

1+x/1!+x^2/2!.................=1+x^2+X…

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Hint: the n-th derivative of asubn x^n evaluated at x = 0 is n!asubn, where asubn is the coefficient of x^n