We know the binomial series

\((1+x)^\alpha=\sum_{n=0}^\infty\left(\begin{array}{c}\alpha\\ n\end{array}\right)x^n\)

Where \(\left(\begin{array}{c}\alpha\\ n\end{array}\right)=\frac{\alpha(\alpha-1)(\alpha-2)...(\alpha-n+1)}{n!}\)

We have given function

\(f(x)=\frac{1}{(1+x)^4}=(1+x)^{-4}\)

By binomial series

\((1+x)^{-4}=\sum_{n=0}^\infty\left(\begin{array}{c}-4\\ n\end{array}\right)x^n\)

\(=\sum_{n=0}^\infty\frac{(-4)(-4-1)(-4-2)...(-4-n+1)}{n!}x^n\)

\(=\sum_{n=0}^\infty\frac{(-4)(-4-1)(-4-2)...(-4-(n-1))}{n!}x^n\)

\(=\sum_{n=0}^\infty\frac{(-1)^n4.5.6.7...(4+(n-1))}{n!}x^n\)

Therefore \(f(x)=\sum_{n=0}^\infty\frac{(-1)^n4.5.6.7...(4+(n-1))}{n!}x^n\)