(a) Since $f_n(x)$ has one change of sign, it follows that $f_n(x) = 0$ has at most one positive real root by Descartes' Rule. As $f_n(0) = -1 < 0$ and $f_n(2) = 2^n - 2^{n-1} - 2^{n-2} - \cdots - 2 - 1 = 1 > 0$, we see that $f_n(x) = 0$ has at least one root between $0$ and $2$. Thus $f_n(x) = 0$ has exactly one root $\alpha_n$ between $0$ and $2$, $n \ge 2$.

(b) We show that $\alpha_n < \alpha_{n+1}$, for $n \ge 2$. Because $f_n(\alpha_n) = 0$, we have $\alpha_n^n - \alpha_n^{n-1} - \alpha_n^{n-2} - \cdots - \alpha_n - 1 = 0$. Now $f_{n+1}(\alpha_n) = \alpha_n(\alpha_n^n - \alpha_n^{n-1} - \alpha_n^{n-2} - \cdots - \alpha_n - 1) - 1 = \alpha_n \cdot 0 - 1 = -1 < 0$. As $f_{n+1}(2) = 1 > 0$, as above, we infer that there is a root of $f_{n+1}(x) = 0$ between $\alpha_n$ and $2$. Since $\alpha_n > 0$, we conclude that $\alpha_n < \alpha_{n+1} < 2$, as desired.

(c) Again
$$(x-1)f_n(x) = (x-1)(x^n - x^{n-1} - x^{n-2} - \cdots - x - 1) = x^{n+1} - 2x^n + 1.$$
So
$$\alpha_n^{n+1} - 2\alpha_n^n + 1 = 0,$$
giving $\alpha_n = 2 - \frac{1}{\alpha_n^n} \ge 2 - \frac{1}{\alpha_2^n} = \frac{1+\sqrt{5}}{2}$, where $\alpha_2 = \frac{1+\sqrt{5}}{2}$, being the positive root of $x^2 - x - 1 = 0$. But $\alpha_2 > 1$. So $\frac{1}{\alpha_2^n} \to 0$, as $n \to \infty$. Thus, in $2 - \frac{1}{\alpha_2^n} \le \alpha_n < 2$, we let $n \to \infty$, to get
$$\lim_{n \to \infty} \alpha_n = 2.$$