On the asymptotics of $a_n=a_{n-1}^k+a_{n-2}^k$ where $k>1$ and $a_0=0, a_1=1$ Consider the sequence defined by $a_0=0,a_1=1, a_n=a_{n-1}^k+a_{n-2}^k$, where $k$ is a fixed integer larger than $1$. 
One finds $a_n\sim a_{n-1}^k$ and thus $a_n\sim \alpha_k^{k^n}$ where $\alpha_k$ is a constant which depends on $k$. 
It seems that $\lim\limits_{k\to\infty}\alpha_k=1$. If so, how can it be proven?

If $k > 1$, it's obvious that the sequence $a_n$ is increasing and nonnegative. So, because $0 \le a_{n-2} \le a_{n-1}$, we have the inequality $a_{n-1}^k \le a_n \le 2a_{n-1}^k$, and that will be all we need to prove the inequality
$$
    2^{k^{n-3}} \le a_n \le 2^{k^{n-1}/(k-1)} \tag{$\star$} \label{eq:star}
$$
for sufficiently large $n$. This doesn't prove that $\alpha_k$ exists (for that we need more precise estimates) but assuming $\alpha_k$ exists, it proves $2^{1/k^3} \le \alpha_k \le 2^{1/(k^2-k)}$, which is enough to know that $\alpha_k \to 1$ as $k \to \infty$ by the squeeze theorem.
(And this is true for various notions of "existence" of $\alpha_k$. For example, we might define $\alpha_k = \lim_{n\to\infty} (a_n)^{1/k^n}$, which leaves open the possibility that $a_n$ only grows like $\alpha_k^{k^n}$ up to some lower-order factors which are "merely exponential", say. Or we might require that $a_n \sim \alpha_k^{k^n}$, which I interpret as saying $\lim_{n\to\infty} \frac{a_n}{\alpha_k^{k^n}} = 1$ and am not yet convinced is true for any $\alpha_k$.)

To prove $\eqref{eq:star}$, let $b_n = \frac{\lg a_n}{k^n}$ (defined for $n\ge 1$). We have
$$
    a_{n-1}^k \le a_n \le 2 a_{n-1}^k \implies k \lg a_{n-1} \le \lg a_n \le k \lg a_{n-1} + 1 \implies b_{n-1} \le b_n \le b_{n-1} + \frac1{k^n}
$$
where the last statement is obtained by dividing through by $k^n$ and applying the definition of $b_n$.
To get a lower bound on $b_n$, just compute the first few values of $a_n$: we have $a_2 = 1^k + 0^k = 1$ and $a_3 = 1^k + 1^k = 2$. So $b_3 = \frac{\lg 2}{k^3} = \frac1{k^3}$, and because $b_{n-1} \le b_n$, we have $b_n \ge \frac1{k^3}$ for all $n \ge 3$.
To get an upper bound on $b_n$, just notice that because $b_{n} \le b_{n-1} + \frac1{k^n}$, then
$$
    b_n \le b_1 + \sum_{i=2}^n \frac1{k^i} \le b_1 + \sum_{i=2}^\infty \frac1{k^i} = \frac1{k(k-1)}.
$$
Knowing that $\frac1{k^3} \le b_n \le \frac1{k(k-1)}$ immediately translates into the bound on $a_n$ in $\eqref{eq:star}$.
