Given that [tex]f^{(n)}(0)=(n+1)![/tex], we have for [tex]f(x)[/tex] the Taylor series expansion about 0 as
[tex]f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n[/tex]
Replace [tex]n+1[/tex] with [tex]n[/tex], so that the series is equivalent to
[tex]f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}[/tex]
and notice that
[tex]\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}[/tex]
Recall that for [tex]|x|<1[/tex], we have
[tex]\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}[/tex]
which means
[tex]f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}[/tex]
[tex]\implies f(x)=\dfrac1{(1-x)^2}[/tex]
