Total probability equal to 1 in Negative Binomial (Pascal) Probability Distribution

Question

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.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}