Looking for analytical proof that this function given as a power series is constant. Answering a recent question I came across the following function ($t$ is a positive integer), defined for $0\le x\le 1$ as:
$$
P_t(x)=x^t\sum_{n=0}^\infty \frac1{n+1}\binom{t(n+1)}n\left[(1-x)x^{t-1}\right]^n.
$$
Is there an analytical way to prove that $P_t(x)$ is a continuous function on $x\in(0,1)$ and:

*

*$P_t(x)=1$ for all $x:\ 0\le 1-x\le\frac1t$,

*$P_t(x)<1$ for all $x:\ \frac1t< 1-x\le1$.

I proved this for trivial cases $t=1,2$ but did not find a way to deal with general $t$.
The behavior of the function for $t=1,2,3,4$ is demonstrated below:

 A: It's analytic about $x=1$. Doing the substitution $x \mapsto 1-x$, collecting coefficients, and using the binomial series quickly reduces it to the following identity. I've used $\tau := t-1$.
Thm: For all $k, \tau \in \mathbb{Z}_{\geq 0}$,
$$\sum_{n=0}^k \frac{(-1)^{k-n}}{n+1} \binom{(n+1)(\tau+1)}{n} \binom{n\tau}{k-n}  = \binom{\tau+k}{k}. \label{*}\tag{*}$$
Proof: This identity was new to me, though @Semiclassical pointed me in the right direction in the comments! As Knuth says in his note on Convolution Polynomials, p.5, we have
$$\sum_{k=0}^n \binom{x+t(n-k)}{n-k} \binom{y+tk}{k} \frac{y}{y+tk} = \binom{x+y+tn}{n}. \label{**}\tag{**}$$
Let $t \mapsto \tau+1$, $n\mapsto k$, $k\mapsto n$, $y\mapsto \tau+1$, and $x\mapsto (\tau+k)-(\tau+1)(k+1)$ in $\eqref{**}$ and applying $\binom{\tau n}{k - n} (-1)^{k-n} = \binom{-\tau n + k - n - 1}{k - n}$ gives $\eqref{*}$. $\Box$
Remark: Knuth notes that $\eqref{**}$ goes back to Rothe in 1793 (!) in the guise of what is apparently known as the Rothe--Hagen identity. Knuth references Gould and Kaucky who provide further references and discussion as well. A short elementary proof is provided by Chu.
A: comment
Interesting.  I tried $n=7$ in Maple, and the result was
$$
P_7(x) = {x}^{7}
{\mbox{$_7$F$_6$}\left(1,{\frac{8}{7}},{\frac{9}{7}},{\frac{10}{7}},{\frac{11}{7}},{\frac{12}{7}},{\frac{13}{7}};\,\frac43,\frac32,\frac53,{\frac{11}{6}},2,{\frac{13}{6}};\,-{\frac { \left( -823543+823543\,x \right) {x}^{6}}{46656}}\right)}
$$
That graph does, indeed, look like yours.  Why?
A: This contribution is a development of the answer given by Joshua and aims mainly at clarification of the remaining questions. It will be assumed that $t>1$.
As already indicated in the mentioned answer it is advantageous to replace $x$ with $1-x$ and consider the function
$$
P_t(x)=(1-x)^tF_t\left(z_t(x)\right)\tag1
$$
with
$$
F_t(z)=\sum_{n=0}^\infty \frac1{n+1}\binom{t(n+1)}n z^n.\tag2
$$
and
$$
z_t(x)=x(1-x)^{t-1}.\tag3
$$
First of all let us investigate the behavior of $z_t(x)$. While $x$ runs from $0$ to $1$ the function first monotonically increases from $0$ to $z^*_t$ and then monotonically decreases back to $0$, the maximum $z^*_t$ being attained at $x=\frac1t$. Thus any value $z:\ 0\le z<z^*_t$ is attained exactly by two values of $x$: $0\le x_1<\frac1t$ and $\frac1t<x_2\le 1$. On the domain $x\in[0,\frac1t]$ the function can be inverted and we shall refer to the inverse function (defined on the domain $z\in[0,z^*_t]$) as $z^{-1}_t(z)$.
For convenience we introduce the function
$$
X_t(x)=z^{-1}_t(z_t(x)).
$$
Observe:
$$
X_t(x)\begin{cases}
=x,& 0\le x\le \frac1t;\\
<x,& \frac1t<x\le 1.
\end{cases}\tag4
$$
The next figure shows the function $z_t(x)$ and indicates the action of the function $X_t(x)$ for $x>\frac1t$.

Let us now investigate the convergence of the series $F_t(z)$ in the neighborhood of $z=0$.
The ratio of the series coefficients reads:
$$\begin{align}
\frac{a_{n}}{a_{n-1}}&=\frac n{n+1}\frac{(nt+t)!}{n!(nt+t-n)!}\frac{(n-1)!(nt-n+1)!}{(nt)!}z\\
&=\frac 1{n+1}\frac{(nt+t)!}{(nt)!}\frac{(n(t-1)+1)!}{(n(t-1)+t)!}z\\
&=\frac {nt+1}{n+1}\frac{nt+2}{n(t-1)+2}\cdot\frac{nt+3}{n(t-1)+3}\cdots\frac{nt+t}{n(t-1)+t}
z,\tag5
\end{align}$$
so that
$$
\lim_{n\to\infty}\frac{a_{n}}{a_{n-1}}=t\left(\frac t{t-1}\right)^{t-1}z.
$$
Thus the series $F_t(z)$ converges if
$$
|z|<\tilde z_t=\frac1t\left(1-\frac1t\right)^{t-1}.\tag6
$$
Observe that $\tilde z_t$ is precisely the maximum value $z^*_t$ of the function $z_t(x)$.
Substituting in (4) $z=z^*_t$ and expanding in powers of $\frac1n$ one obtains
$$
\frac{a_{n}}{a_{n-1}}=1-\frac32\frac1n
+O\left(\frac1{n^2}\right),
$$
so that by Raabe test:
$$
\lim_{n\to\infty}n\left(\frac{a_{n-1}}{a_{n}}-1\right)=\frac32>1
$$
the series converges also for $z=z^*_t$.
After the absolute convergence is established one can demonstrate applying the Rothe-Hagen identity:
$$\forall x: 0\le x\le\frac1t:\ F_t\left(z_t(x)\right)=\frac1{(1-x)^t}.\tag7
$$
The proof is given in the answer of Joshua and need not be repeated.
Using (7) one obtains:
$$\begin{align}
P_t(x)&=(1-x)^tF_t(z_t(x))\\
&=\frac{(1-x)^t}{(1-X_t(x))^t}
\underbrace{\left[(1-X_t(x))^t F_t(z_t(X_t(x)))\right]}_{=1}\\
&=\left(\frac{1-x}{1-X_t(x)}\right)^t.\tag8
\end{align}$$
In view of the property (4) one concludes that indeed:
$$
P_t(x)\begin{cases}
=1,& 0\le x\le \frac1t;\\
<1,& \frac1t<x\le 1.
\end{cases}\tag9
$$
A: Too long for comments.
Assuming that $t$ is a positive integer, there is no problem to get explicit formulae up to $t=3$.
For $t \geq 4$ appear hypergeometric function and the result write
$$P_t(x)=\frac x {1-x} \big[Q_t(x)-1\big]$$
$$Q_4(x)=\,
   _3F_2\left(\frac{1}{4},\frac{2}{4},\frac{3}{4};\frac{2}{3},\frac{4}{3};\frac{4^4}{3^3} (1-x) x^3\right)$$
$$Q_5(x)=\,
   _4F_3\left(\frac{1}{5},\frac{2}{5},\frac{3}{5},\frac{4}{5};\frac{2}{4},\frac{3}{4},\frac{5}{4};\frac{5^5}{4^4} (1-x) x^4\right)$$
$$Q_6(x)=\,
   _5F_4\left(\frac{1}{6},\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6};\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{6}{5};\frac{6^6} {5^5}(1-x) x^5\right)$$
$$Q_7(x)=\,
   _6F_5\left(\frac{1}{7},\frac{2}{7},\frac{3}{7},\frac{4}{7},\frac{5}{7},\frac{6}{7};\frac{2}{6},\frac{3}{6},\frac{4}{6},\frac{5}{6},\frac{7}{6};\frac{7^7}{6^6} (1-x) x^6\right)$$
For $x=x_*=1-\frac 1t$, $Q_t(x_*)=\frac t{t-1}$ and then $P_t(x_*)=1$ as you observed.
