Binomial expansion of $(1-x)^n$

Well, as I understand it, we could write the binomial expansion as:

$$(1-x)^n= \sum_{k=0}^{n} \binom n k 1^{n-k}\,(-x)^k$$ $$\binom{n}{0}1^n (-x)^0 + \binom n 1 1^{n-1} (-x)+ \binom n 2 1^{n-2}(-x)^2 + \binom n 3 1^{n-3}(-x)^3 \ldots$$

which simplifies to $$1-nx+\frac{n(n-1)}{2!}\cdot x^2 -\frac{n(n-1)(n-2)}{3!} \cdot x^3 \ldots$$

Which is the answer everyone else has given.

It is

$= 1+ n(-x) + (n(n-1)(-x)^2) / 2! + (n(n-1)(n-2)(-x)^3) / 3!$

I'm not sure how appropriate it is to answer questions this old, but compared to the methods above, I feel the easiest way to see the answer to this question is to take

a = -x

And substitute that into the binomial expansion:

(1+a)^n

This yields exactly the ordinary expansion. Then, by substituting -x for a, we see that the solution is simply the ordinary binomial expansion with alternating signs, just as everyone else has suggested.

In my opinion, this substitution is the best way to see "how" to get the binomial expansion, as the OP originally asked, because it demonstrates a method which reduces the problem to the expression OP already has, but shows how one can eliminate the added complexity of the minus sign, and explicitly justifies the treatment of -x used in the other answers.