Total probability equal to 1 in Negative Binomial (Pascal) Probability Distribution Prove that $\displaystyle \sum^\infty_{t=k} \binom{t-1}{k-1}p^{k}(1-p)^{t-k}$ sum converges to $1$ while $p\in[1,0]$. $$ $$
I tried $$\sum^\infty_{t=k} \binom{t-1}{k-1}p^{k}(1-p)^{t-k}\\=\frac{p^k}{(k-1)!}\sum^\infty_{t=k} \left((1-p)^{t-1}\right)^{(k-1)}\\=\frac{p^k}{(k-1)!}\left(\sum^\infty_{t=k} (1-p)^{t-1}\right)^{(k-1)}\\=\frac{p^k}{(k-1)!}\left(\frac{(1-p)^{k-1}}{p}\right)^{(k-1)}$$
It is where I am stuck. If there was not $(1-p)^{k-1}$, I would be able to prove it easily but there is. I am sure the way is right, I made a mistake in calculations but I can not find it. Sorry for my insufficient language.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\sum_{t = k}^{\infty}{t - 1 \choose k - 1}p^{k}\pars{1 - p}^{t - k} =
p^{k}\sum_{t = 0}^{\infty}{t + k - 1 \choose k - 1}\pars{1 - p}^{t}
\\[5mm] = & \
p^{k}\sum_{t = 0}^{\infty}{t + k - 1 \choose t}\pars{1 - p}^{t}
=
p^{k}\sum_{t = 0}^{\infty}{-k \choose t}\pars{-1}^{t}\pars{1 - p}^{t}
\\[5mm] = & \
p^{k}\sum_{t = 0}^{\infty}{-k \choose t}\pars{p - 1}^{t}
=
p^{k}\,\bracks{\vphantom{\large A}1 + \pars{p - 1}}^{\,-k} = \bbx{1} \\ &
\end{align}
