Trouble with the Binomial sum I guess the following sum is equal to $1$ since you can compute the baby cases for $t=1,2$ straightforward to get the desired result. So, in general, we should have the following sum to be $1$, I think. But, I am stuck in the last equality. Any comments or advice would be appreciated.
\begin{align*}
    &\frac{t}{n}+t\left(\frac{\binom{n-t}{1}}{2\binom{n}{2}}+\frac{\binom{n-t}{2}}{3\binom{n}{3}}+\frac{\binom{n-t}{3}}{4\binom{n}{4}}+\cdots+\frac{\binom{n-t}{n-t}}{(n-t+1)\binom{n}{n-t+1}}\right)\\\\
    =~~&\frac{t}{n}+t\left(\frac{\binom{n-t}{1}}{n\binom{n-1}{1}}+\frac{\binom{n-t}{2}}{n\binom{n-1}{2}}+\frac{\binom{n-t}{3}}{n\binom{n-1}{3}}+\cdots+\frac{\binom{n-t}{n-t}}{n\binom{n-1}{n-t}}\right)\\\\
    =~~&\frac{t}{n}+\frac{t}{n}\left(\frac{\binom{n-t}{1}}{\binom{n-1}{1}}+\frac{\binom{n-t}{2}}{\binom{n-1}{2}}+\frac{\binom{n-t}{3}}{\binom{n-1}{3}}+\cdots+\frac{\binom{n-t}{n-t}}{\binom{n-1}{n-t}}\right)
\end{align*}
where the first equality is used $k\binom{n}{k}=n\binom{n-1}{k-1}$ and $t\in\{1,2,...,n\}.$

$$\frac tn\sum_{i=0}^{n-t}\frac{\binom{n-t}i}{\binom{n-1}i}=\frac1{\binom nt}\sum_{i=0}^{n-t}\binom{n-1-i}{t-1}=\frac1{\binom nt}\sum_{j=0}^{n-t}\binom{t-1+j}{t-1}=1.$$
The first equality is obtain by using factorials, the second one by letting $j=n-t-i,$ and the last one is the hockey-stick identity.
