General expression of polynomial roots sequence. Let $$P_n(X) = X^n-X^{n-1}-\cdots-1 = X^n - \sum_{k=0}^{n-1} X^k$$
The question is : Find the general expression of the sequence $(u_n)_{n \in \mathbb{N}}$ where $u_k$ is the greater root of $P_{k+1}(X)$.
We aldready know that $$u_0 = 1$$, $$u_1 = \frac{1+\sqrt{5}}{2}$$ and $$u_2 = \frac{1}{3}+\frac{1}{3} \sqrt[3]{19-3\sqrt{33}}+\frac{1}{3}\sqrt[3]{19+3\sqrt{33}}$$
We also can also think (by observations) that $u_h \to 2$ when $h \to \infty$.
But I can't find the general expression for $u_n$.

By Descartes rule of signs, $P_n(x)$ has one sign change, so it has one root on the positive half of the real line.  $P_n(2) = 1$ and $P_n \left( 2-\frac{1}{n} \right) = \frac{n - \left( 2-\frac{1}{n} \right)^n}{n-1}$.  Since $n-1>0$ for $n > 1$, the sign of this latter expression is controlled by the numerator.  If we can show the numerator is negative for $n$ sufficiently large, the root must lie in $(2-\frac{1}{n}, 2)$.  As the lower end of this interval is approaching $2$, the limit of the only positive root of $P_n$ approaches $2$ as $n \rightarrow \infty$.
For $n > 2$, $2 - \frac{1}{n} > 3/2$, so $\left( 2 - \frac{1}{n} \right)^n > \left( \frac{3}{2} \right)^n$.  We find that $\frac{\mathrm{d}}{\mathrm{d}n} \left( n - \left( \frac{3}{2} \right)^n \right) = 0 $ when $n = \frac{\ln(\ln 3 - \ln 2)}{\ln 2 - \ln 3}$ at which, $\frac{\ln(\ln 3 - \ln 2)}{\ln 2 - \ln 3} - \left( \frac{3}{2} \right)^{\frac{\ln(\ln 3 - \ln 2)}{\ln 2 - \ln 3}} = \frac{-(1+\ln\ln(3/2))}{\ln(3/2)} = -0.2399{\dots} < 0$.  Therefore, the numerator above is negative (at least) for all $n > 2$.
