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?
 A: 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 $\eqref{eq:star}$.
A: Misha showed that the limit in the OP is $1$. I'll prove in particular $\lim\limits_{k\to\infty} \alpha_k^{k^3}=\lim\limits_{n\to\infty}a_n^{1/k^{n-3}}=2$.
On one hand, by induction it's easy to prove $a_n>2^{k^{n-3}}$, and it does come to mind when looking at the first terms. On the other hand, we find $a_n<2^{k^{n-3}+k^{n-5}(k+n-5)}$ for $n>5$. This is a (loose) upper bound that, again, one may think of while manipulating the first terms. For example $a_4=2^k+1, a_5=2^{k^2}\left(\left(1+2^{-k}\right)^k+1\right)$ and \begin{align}a_6<(a_5+a_4)^k &=2^{k^3}\left(\left(1+2^{-k}\right)^k+2^{k-k^2}+2^{-k^2}\right)^k \\ &< 2^{k^3}\left(2^k+2\right)^k<2^{k^3+k(k+1)}.\end{align}Analogously it holds for $a_7$ as well; assuming it for $a_{n-1}$ and $a_{n-2}$ it holds for $a_n$, $n\ge8$, since then \begin{align} a_n &< 2^{k^{n-3}}\left(2^{k^{n-6}(k+n-6)}+2^{k^{n-5}-k^{n-4}+k^{n-7}(k+n-7)}\right)^k \\ &< 2^{k^{n-3}}\left(2^{k^{n-6}(k+n-6)}+2^{2^{n-5}-2^{n-4}+2^{n-7}(n-5)}\right)^k \\ &= 2^{k^{n-3}}\left(2^{k^{n-6}(k+n-6)}+2^{2^{n-7}(n-9)}\right)^k \\ &< 2^{k^{n-3}}\left(2^{k^{n-6}(k+n-6)+1}\right)^k \\ &< 2^{k^{n-3}+k^{n-5}(k+n-5)}\end{align} where we have used the decreasingness of $x\mapsto x^{n-5}-x^{n-4}+x^{n-7}(x+n-7)$ over $[2,\infty)$, the equality $2^{n-5}-2^{n-4}+2^{n-7}(n-5)=2^{n-7}(n-9)$ and the inequality $k^{n-6}(k+n-6)\ge 2^{n-6}(n-4)>2^{n-7}(n-9)$.
