Gosper's Identity $\sum_{k=0}^n{n+k\choose k}[x^{n+1}(1-x)^k+(1-x)^{n+1}x^k]=1 $

The page on Binomial Sums in Wolfram Mathworld http://mathworld.wolfram.com/BinomialSums.html (Equation 69) gives this neat-looking identity due to Gosper (1972):

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

Would anyone know if there is a simple proof of this identity without using induction?

Hint: Suppose $0 \leq x \leq 1$, and consider a coin with bias $x$ being flipped until one of its sides comes up $n+1$ times. The left-hand side counts the probability that this happens, which is plainly $1$. Since both sides are polynomials in $x$ and the identity is true for infinitely many values of $x$, it must be true for all $x$.

Suppose we seek to show that $$\sum_{k=0}^n {n+k\choose k}(x^{n+1}(1-x)^k+(1-x)^{n+1}x^k) =1.$$

We use an Iverson bracket to control the range of $k$ so we can let it range from zero to infinity, which is

$$[[0 \le k\le n]] = \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{1+v+\cdots+v^n}{v^{k+1}} \; dv \\ = \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{v^{n+1}-1}{(v-1)v^{k+1}} \; dv.$$

We evaluate this using the formula for the residue at infinity $$\mathrm{Res}_{z=\infty} h(z) = \mathrm{Res}_{z=0} \left[-\frac{1}{z^2} h\left(\frac{1}{z}\right)\right]$$

which in this case yields (omit the minus sign as the residues sum to zero) $$\mathrm{Res}_{v=0} \frac{1}{v^2} \frac{1/v^{n+1}-1}{(1/v-1) \times 1/v^{k+1}} = \mathrm{Res}_{v=0} \frac{1/v^{n+1}-1}{(1-v) \times 1/v^{k}} \\ = \mathrm{Res}_{v=0} \left(\frac{v^{k}}{v^{n+1}}\frac{1}{1-v} -\frac{v^k}{1-v}\right).$$

With the additional assumption that $k \ge 0$ which is the case here this yields $$\mathrm{Res}_{v=0} \frac{v^{k}}{v^{n+1}}\frac{1}{1-v} = \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{v^{k}}{v^{n+1}}\frac{1}{1-v} \; dv$$ which could have been obtained by inspection.

This yields for the second component of the sum $$\frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{1}{v^{n+1}}\frac{1}{1-v} \sum_{k\ge 0} {n+k\choose n} (1-x)^{n+1} x^{k} v^k \; dv \\ = (1-x)^{n+1} \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{1}{v^{n+1}}\frac{1}{1-v} \frac{1}{(1-xv)^{n+1}} \; dv.$$

We will evaluate this not by evaluating the residue at zero but the sum of the negatives of the residues at $v=1$ and $v=1/x$ given that the residues sum to zero.

For the residue at $v=1$ re-write the integral as follows: $$-(1-x)^{n+1} \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{1}{v^{n+1}}\frac{1}{v-1} \frac{1}{(1-xv)^{n+1}} \; dv.$$

The residue at $v=1$ here is $$-(1-x)^{n+1} \frac{1}{(1-x)^{n+1}} = -1$$ for a contribution of $$1$$ upon negation.

For the residue at $v=1/x$ re-write the integral as follows: $$\frac{(1-x)^{n+1}}{x^{n+1}} \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{1}{v^{n+1}}\frac{1}{1-v} \frac{1}{(1/x-v)^{n+1}} \; dv \\ = (-1)^{n+1} \frac{(1-x)^{n+1}}{x^{n+1}} \frac{1}{2\pi i} \int_{|v|=\epsilon} \frac{1}{v^{n+1}}\frac{1}{1-v} \frac{1}{(v-1/x)^{n+1}} \; dv.$$

Use Leibniz' rule to differentiate the two terms in $v$ to get

$$\frac{1}{n!}\left( \frac{1}{v^{n+1}}\frac{1}{1-v} \right)^{(n)} = \frac{1}{n!} \sum_{k=0}^n {n\choose k} \frac{(-1)^k (n+k)!}{n! \times v^{n+1+k}} \frac{(n-k)!}{(1-v)^{n-k+1}}.$$

Evaluate this at $v=1/x$ including the factor in front to get for the residue $$(-1)^{n+1} \frac{(1-x)^{n+1}}{x^{n+1}} \frac{1}{n!} \sum_{k=0}^n {n\choose k} \frac{(-1)^k (n+k)!}{n! \times (1/x)^{n+1+k}} \frac{(n-k)!}{(1-1/x)^{n-k+1}} \\ = (-1)^{n+1} \frac{(1-x)^{n+1}}{x^{n+1}} \sum_{k=0}^n \frac{(-1)^k (n+k)!}{n! \times k! \times (1/x)^{n+1+k}} \frac{1}{(1-1/x)^{n-k+1}} \\ = (-1)^{n+1} (1-x)^{n+1} \sum_{k=0}^n {n+k\choose k} \frac{(-1)^k}{(1/x)^{k}} \frac{1}{(x-1)^{n-k+1}/x^{n-k+1}} \\ = \sum_{k=0}^n {n+k\choose k} \frac{(-1)^k}{(1/x)^{k}} \frac{1}{(x-1)^{-k}/x^{n-k+1}} \\ = x^{n+1} \sum_{k=0}^n {n+k\choose k} (1-x)^k.$$

Upon negation this becomes the negative of the first component of the sum. Hence adding the three pieces (first component, one, negative of first component) we obtain a sum of

$$1.$$

Remark. If we want to do this properly we also need to verify that the residue at infinity of the integral in $v$ is zero.

In the present case this becomes

$$- \mathrm{Res}_{v=0} \frac{1}{v^2} \frac{1}{(1/v)^{n+1}}\frac{1}{1-1/v} \frac{1}{(1-x/v)^{n+1}} \\ = - \mathrm{Res}_{v=0} \frac{1}{v^2} \frac{v^{n+1} \times v^{n+1}}{1-1/v} \frac{1}{(v-x)^{n+1}} \\ = - \mathrm{Res}_{v=0} \frac{1}{v} \frac{v^{2n+2}}{v-1} \frac{1}{(v-x)^{n+1}} \\ = - \mathrm{Res}_{v=0} \frac{v^{2n+1}}{v-1} \frac{1}{(v-x)^{n+1}}$$ which is zero by inspection.