Looking for analytical proof that this function given as a power series is constant.
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.
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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?