Think outside the box

$\begin{aligned} f^{(0)} (x) & = (1+2x-x^2)^4 \\ & = (2-[x^2-2x+1])^4 \\ & = (2-[x-1]^2)^4 \\ & = 2^4-4(2^3)(x-1)^2 + 6(2^2)(x-1)^4 -4(2)(x-1)^6 +(x-1)^8 \\ \implies f^{(7)} (x) & = \frac {d^7}{dx^7} 2^4-\frac {d^7}{dx^7}4(2^3)(x-1)^2 + \frac {d^7}{dx^7} 6(2^2)(x-1)^4 - \frac {d^7}{dx^7} 4(2)(x-1)^6 + \frac {d^7}{dx^7} (x-1)^8 \\ & = 0 -0+0-0 + 8!(x-1)^8 \\ f^{(7)} (0) & = -8! \\ \implies k & = \boxed{8} \end{aligned}$

consider the taylor series expansion of the function.

$f^{(0)}(x) = (1+2x-x^2)^4$

here,the coefficient of $x^{7}$ is $\frac{f^{(7)}(0)}{7!}$

now the coefficient of $x^{7}$ in the function given is clearly $-8$ thus

$\frac{f^{(7)}(0)}{7!}$ is $-8$

so $f^{(7)}(0)$ is $-8!$

$f(x)=(1+2x-x^2)^4 \\ =(2-1+2x-x^2)^4 \\ =(2-(1-x)^2)^4$

After expanding this there will be 4 terms and a constant term.The four terms has powers of $(x-1)$ $8,6,4,2$ . After differentiating seven times all terms will be zero except $(x-1)^8$ .

$f^7(x)=8!(x-1)$ .

So, $f^7(0)=-8!$