<s>[INST] 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?
 [/INST] Your proof can be slightly modified as follows:
Note that 
$x(x+1)^{n-1}=x^n+{n-1 \choose 1}x^{n-1}+{n-1 \choose 2}x^{n-2}+...+{n-1 \choose n-2}x^2+x$
$(x+1)^{n-1}=x^{n-1}+{n-1 \choose 1}x^{n-2}+{n-1 \choose 2}x^{n-3}+...+{n-1 \choose n-2}x+1$
Add the 2 equations and look at the addition of like terms.
$(x+1)^n=x^n + ({n-1 \choose 1}+{n-1 \choose 0})x^{n-1} + ({n-1 \choose 2}+{n-1 \choose 1})x^{n-2}+...+({n-1 \choose n-1}+{n-1 \choose n-2})x+1$
Expand $(x+1)^n$ and see how the identity can be proved.
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