We have to distinguish two cases:

(1) If # p in NN# then the derivatives of order grater than #p# of #f(x)# are identically null. In such case:

#(df)/dx = px^(p-1)#

...

#(df^k)/dx^k = p(p-1)...(p-k+1)x^(p-k) = (p!)/(p-k!)x^(p-k)#

and for #x=1#

#[(df^k)/dx^k ]_(x=1) = (p!)/(p-k!)#

so the Taylor series has a finite number of terms:

#x^p = sum_(k=0)^p (p!)/(p-k!)(x-1)^k/(k!) = sum_(k=0)^p ((p),(k)) (x-1)^k#

where #((p),(k)) = (p!)/((k!)(p-k)!)# is a binomial coefficient.

(2) If # p notin NN# then we can use the definition of generalized binomial coefficient:

#((p),(k)) = (p(p-1)(p-2)...(p-k+1))/(k!)#

where #p in RR# and #k in NN#

In this case the derivatives of all orders are:

#(df^k)/dx^k = p(p-1)...(p-k+1)x^(p-k) = ((p),(k)) k! x^(p-k)#

and for #x=1#:

#[(df^k)/dx^k ]_(x=1) = ((p),(k)) (k!)#

so the Taylor series is:

#x^p = sum_(k=0)^oo ((p),(k)) (k!)(x-1)^k/(k!) = sum_(k=0)^oo ((p),(k)) (x-1)^k#

Note that if we use the definition of generalized binomial coefficient, we have:

#((p),(k)) = (p(p-1)(p-2)...color(blue)((p-p))..(p-k))/(k!) = 0#

for #p,k in NN# and #k > p#

then we can unify the notation as:

#x^p = sum_(k=0)^oo ((p),(k)) (x-1)^k#

for any #p#.