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?
 A: 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.
A: A much shorter way would be calculating the coefficient of $x^k$ independently from each side of $$(1+x)^n=(1+x)(1+x)^{n-1}$$
Using the binomial theorem, the coefficient of $x^k$ on the LHS is $\binom{n}{k}$.
For the right hand side, $x^k$ can be formed by two ways -


*

*Selecting $1$ from $(1+x)$ and $x^k$ from $(1+x)^{n-1}$. This gives coefficient $\binom{n-1}{k}$.

*Selecting $x$ from $(1+x)$ and $x^{k-1}$ from $(1+x)^{n-1}$. This gives coefficient $\binom{n-1}{k-1}$.


So the total coefficient of $x^k$ on the RHS is $\binom{n-1}{k} + \binom{n-1}{k-1}$.
Since LHS=RHS, $$\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$$
A: 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.
