Density of heads with a biased coin stopped after k heads

If a biased coin (prob of heads=p) is flipped until you get $k$ heads, and $X$ is the number of flips, you can compute that the expected value of $(k-1)/(X-1)$ is p by computing the sum $\mathbb{E}\left[\frac{k-1}{X-1}\right] = (k-1)\sum_{n=k}^\infty \frac{1}{n-1} \binom{n-1}{k-1} p^k (1-p)^{n-k}=p.$ It seems like there should be a more philosophical answer though. The last flip in the sequence is always heads, so this is saying that we expect the density of heads in the rest of the sequence to be the bias of the coin. Is there a more satisfying / philosophical / intuitive solution?