How can the derivative be used with this:
$$\sum_{k=0}^n {n \choose k}x^k = (1+x)^n$$
to generate a second identity?
In a recent question, one of the answers involved taking the derivative of the binomial theorem to get to a second identity.
It is not clear to me how one is able to get from:
$$\sum_{k=0}^n {n \choose k}x^k = (1+x)^n$$
to some intermediate form which when $x = -1, n=5$ gives:
$$\sum_{k=0}^{5}k^4{5\choose k}(-1)^k = 0$$
On
If you took the derivative of both sides and then multiplied both sides by $x$ you would get
$$\sum_{k=0}^n k{n \choose k}x^{k-1} = n(1+x)^{n-1}$$ $$\sum_{k=0}^n k{n \choose k}x^{k} = nx(1+x)^{n-1}$$
Then you would repeat that three more times so that you would get the $k^4$ factor on the LHS. I'm not going to do the Product Rule on a Saturday for you, but whatever it is on the RHS it would equal $0$ when $x=-1$ if $n$ was sufficiently large (i.e. $n>4$) because there would be a (1+x) factor in every term.
Take the equality for $n=5$ and differentiate it $4$ times giving $$\sum_{k=0}^5\binom5kx^k=(1+x)^5$$ $$\sum_{k=0}^5k\binom5kx^{k-1}=5(1+x)^4$$ $$\sum_{k=0}^5k(k-1)\binom5kx^{k-2}=20(1+x)^3$$ $$\sum_{k=0}^5k(k-1)(k-2)\binom5kx^{k-3}=60(1+x)^2$$ $$\sum_{k=0}^5k(k-1)(k-2)(k-3)\binom5kx^{k-4}=120(1+x)$$ Then multiply each identity by $x$ until we have $x^k$ in the summation thus giving $$\sum_{k=0}^5\binom5kx^k=(1+x)^5$$ $$\sum_{k=0}^5k\binom5kx^k=5x(1+x)^4$$ $$\sum_{k=0}^5k(k-1)\binom5kx^k=20x^2(1+x)^3$$ $$\sum_{k=0}^5k(k-1)(k-2)\binom5kx^k=60x^3(1+x)^2$$ $$\sum_{k=0}^5k(k-1)(k-2)(k-3)\binom5kx^k=120x^4(1+x)$$ Then note that $$k^4=k(k-1)(k-2)(k-3)+6k(k-1)(k-2)+7k(k-1)+k$$ hence we can combine the identities to give $$\sum_{k=0}^5k^4\binom5kx^k=120x^4(1+x)+6\times60x^3(1+x)^2+7\times20x^2(1+x)^3+5x(1+x)^4$$ Then using $x=-1$ gives the identity $$\sum_{k=0}^5k^4\binom5k(-1)^k=0$$
Note that it is always possible to write $$k^n=a_0k(k-1)\cdots(k-n+1)+a_1k(k-1)\cdots(k-n+2)+\dots+a_{n-1}k$$ for some constants $a_i\in\mathbb{Z}$.