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?

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 $k^{n-6}(k+n-6)\ge 2^{n-6}(n-4)>2^{n-7}(n-9)$.
