Estimation of a geometric distribution's parameter by the reciprocal of the sample mean

I came accross this exercise when studying statistics, but I can't get to what's the solution. The exercise simply asks to show whether the reciprocal of the sample mean is an unbiased estimation for the unknown parameter $$p$$ if the sample is taken from the distribution $$Geo(p)$$.

I've tried the following: Let $$\underline{x}$$ be our sample, with size $$n$$. Let our estimation be $$T(\underline{x}) = \frac{n}{\sum_{i=1}^{n} x_i}$$. We need to show that $$E(T(\underline{x})) = p$$.

The sum of $$n$$ geometric distributions follows a $$NegBinom(n, p)$$ distribution, so with this, if I'm not mistaken the mean of our estimation goes like this:

$$E(T(\underline{x})) = n\sum_{i=n}^{\infty}\frac{1}{i}\binom{i}{n}(1-p)^{i-n}p^n = \sum_{i=n}^{\infty}\binom{i-1}{n-1}p^n(1-p)^{i-n}$$

Since our estimation has to be equal $$p$$, I arranged the two sides as: $$\sum_{i=n}^{\infty}\binom{i-1}{n-1}p^n(1-p)^{i-n} = p$$. From this, $$\sum_{i=n}^{\infty}\binom{i-1}{n-1}p^{n-1}(1-p)^{i-n} = 1$$ should be shown (with $$p \neq 0$$). It obviously reminds of a binomial series, but I can't progress from this spot. Any help would be appricated.

By Jensen inequality, $$\mathbb E\left(\frac{n}{\sum_{i=1}^{n} x_i}\right)>\frac{n}{\mathbb E(\sum_{i=1}^{n} x_i)}=p$$ So this estimator cannot be unbiased.
The exact value is $$\mathbb E(T(\underline{x})) = n\sum_{i=n}^{\infty}\frac{1}{i}\binom{i-1}{n-1}(1-p)^{i-n}p^n$$ and this value leads to a hypergeometric function and does not have a simple form.
Form the other side, if you consider $$\frac{n}{\sum_{i=1}^n x_i -1}$$ instead of $$T(\underline x)$$, then the expected value can be found: $$\mathbb E\left(\frac{n}{\sum_{i=1}^n x_i -1}\right) = n\sum_{i=n}^{\infty}\frac{1}{i-1}\binom{i-1}{n-1}(1-p)^{i-n}p^n$$ $$= \frac{n}{n-1}p \sum_{i=n}^\infty \binom{i-2}{n-2}p^{n-1}(1-p)^{i-n} = \frac{n}{n-1}p,$$ since the last sum equals to $$1$$ as the sum of probabilities of all possible values of $$NegBinom (n-1,p)$$ distribution: $$1=\sum_{j=n-1}^\infty \binom{j-1}{n-2}p^{n-1}(1-p)^{j-n+1}=\bigl[i=j+1\bigr]=\sum_{i=n}^\infty \binom{i-2}{n-2}p^{n-1}(1-p)^{i-n}.$$ And from $$\mathbb E\left(\frac{n}{{\sum_{i=1}^n x_i} -1}\right) = \frac{n}{n-1}p$$ you can design unbiased estimate if needed.