# Is this a good proof of the binomial identity?

Prove that the binomial identity ${n\choose k} = {n-1\choose k-1} + {n-1\choose k}$ is true using the following expression: $(1+x)^n = (1+x)(1+x)^{n−1}$ and the binomial theorem.

What I have:

We know from the binomial theorem that:

$$(x+1)^n= {n\choose 0} x^0 + {n\choose 1} x^1+\cdots+{n\choose k} x^k+\cdots+ {n\choose n-1} x^{n-1}+ {n\choose n} x^n.$$

By using the property that $(1+x)^{n} = (1+x)(1+x)^{n−1}$, we can take out a factor of $x$ to get:

$$x(x+1)^{n-1}={n-1\choose 0} x^0+\cdots+{n-1\choose k-1} x^{k-1}+\cdots+{n+1\choose n-1} x^{n-1}$$

If we then divide by a factor of $x$ we get:

$$1(x+1)^{n-1}={n-1\choose 0} x^0+\cdots +{n-1\choose k-1} x^k+\cdots+{n-1\choose n-1} x^{n-1}$$

Substituting $(x+1)^{n-1}={n-1\choose 0} x^0 +\cdots+{n-1\choose k-1} x^k+\cdots+{n-1\choose n-1} x^{n-1}$ into the equation $x(x+1)^{n-1}= {n-1\choose 0} x^0+\cdots+{n-1\choose k-1} x^{k-1}+\cdots+ {n+1\choose n-1}x^{n-1}$, the equation can be reduced to:

$${n\choose k}x^k = {n-1\choose k-1} x^{k-1}x+{n-1 \choose k}x^k 1$$

Dividing by $x$ we get: ${n\choose k} = {n-1\choose k-1} + {n-1\choose k}$

Is this a good proof? What can I do to improve it? Is there a better way to solve this problem?