If $f(x)=x^n$, then the value of $f(1)-\frac{f^{\prime }(1)}{1!}+\frac{f^{\prime \prime }(1)}{2!}-\frac{f^{\prime \prime \prime }(1)}{3!}+\cdots +(-1{)}^n\frac{f^n(1)}{n!}$ is:

The expression is the Taylor series expansion of $f(x)$ centered at $a=1$, evaluated at some point.Specifically, let's look at the terms:$f(1)=1$$f^{\prime }(1)=n$$f^{\prime \prime }(1)=n(n-1)$ The expression is:
$$1-\frac{n}{1!}+\frac{n(n-1)}{2!}-\cdots +(-1{)}^n\frac{n!}{n!}$$
This is exactly the binomial expansion of $(1-1{)}^n$.
$$(1-1{)}^n=\sum _{r=0}^n{}^n{C}_r(-1{)}^r=1-n+\frac{n(n-1)}{2}-\cdots =0$$
Thus the value is 0.