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?

First of all, in general
$$x(x+1)^{n-1} \neq \binom{n-1}{0} x^0+\cdots+\binom{n-1}{k - 1}x^{k-1}+\cdots
+ \binom{n+1}{n-1}x^{n-1}.$$
Notice that as $x\to 0$ the left side goes to zero but the right side does not.
Also, in general
$$1(x+1)^{n-1} \neq \binom{n-1}{0}x^0+\cdots+\binom{n-1}{k - 1}x^{k}+\cdots+
\binom{n-1}{n - 1}x^{n-1}.$$
If you change $x^k$ to $x^{k-1}$ on the right hand side, the two sides are equal.
The step where you "reduce" an equation to
$\binom nk x^{k} = \binom{n-1}{k - 1} x^{k-1}x + \binom{n-1}{k}x^{k}1$
has no clear explanation, but it doesn't really matter since the
premises of that statement are already false.
