Linear asymptotics for the solution to $\sum_{n=0}^{\infty} \left(n^k -\frac{1}{\alpha}\right) c^{n^k}=0$ Consider the function 
$f_k(c):=\sum_{n=0}^{\infty} c^{n^k}$ where $k\ge 1$ is an integer. This one obviously converges for $\left\lvert c \right\rvert <1.$
In the following we study the solution to the equation 
$$\sum_{n=0}^{\infty} \left(n^k -\frac{1}{\alpha}\right) c^{n^k}=0.$$
This one always exists as long as $\alpha \in (0,1).$
Numerically, I discovered something that I would like to understand:
As $\alpha \rightarrow 0$ we have that $c= 1-\frac{\gamma\alpha}{k}$ for some constant $\gamma.$
So first the solution $c$ seems to depend in a linear way on $\alpha$ for $\alpha$ small and second, the dependence on $k$ also seems to be just $1/k$. 
I would like to understand these two observations.  
 A: We can indeed show that for all $k$,
$$1-c \sim_k \frac{\alpha}{k} \qquad (\alpha \to 0^+)$$
The middle part of my answer makes it intuitive why the main term is linear in $\alpha$. What it does not explain, is whether there is a deeper reason that it is linear in $1/k$.
Uniqueness of the solution. Let's start by justifying that there is a unique solution $c \in (0,1)$ for small $\alpha$.
We have $$\alpha = \frac{f_k(c)}{c \cdot f_k'(c)}$$
so we see that $\alpha$, as a function of $c$, has a pole at $0$ of order $1$. In particular, $\alpha(c) \to +\infty$ as $c \to 0^+$. 
In the sequel, we always assume $c \in (0,1)$, and often write simply $f_k$ for $f_k(c)$. For the derivative we have $$\alpha'(c) = \frac{c(f_k')^2 - f_k(cf_k')'}{(cf_k')^2}$$
By Cauchy-Schwarz, $(f_k')^2 < c f_k(cf_k')'$ so that $\alpha$ is strictly decreasing.
We have $\alpha \to 0$ for $c \to 1^-$. Indeed, this follows e.g. by A Tauberian theorem for a quotient of power series, the limit on the boundary
There is probably a more elementary argument, but we will use the same result again anyway.
Using $\alpha'<0$, $\alpha > 0$, $\alpha \to +\infty$ for $c \to 0^+$ and $\alpha \to 0$ for $c \to 1^-$, we conclude that $\alpha : (0,1) \to (0, \infty)$ is a decreasing bijection.
Asymptotics. First take $k=1$, so that $f_k(c) = (1-c)^{-1}$ and we simply have $$\alpha = \frac1c-1$$
Let $k \geq 1$ be arbitrary now. We will determine the limit of $\alpha'$ as $c \to 1$, but first finish the argument. Suppose the limit exists and equals $p_k \neq 0$. Then $\alpha$ has a $C^1$-extension to the right of the point $c=1$, and we have have $$\alpha(c) = p_k \cdot (c-1) + o_k(1) \qquad (\alpha \to 0, c \to 1^-)$$
The statement from the beginning of the post follows, with $p_k=-k$.
Limit of the derivative $\alpha'(c)$.
We have
$$\alpha' = 1- \frac{f(c)}{c g(c)}$$
with $$f(c) := cf_k(cf_k')' = \sum_{m,n \geq 0}c^{m^k+n^k}n^{2k} =: \sum_{i \geq 0}a_ic^i$$
and $$g(c) := (cf_k')^2 = \sum_{m,n\geq 0}c^{m^k+n^k}m^kn^{k} =: \sum_{i \geq 0}b_ic^i$$
By the linked question, we only have to estimate the partial sums of the coefficients $a_n$ and $b_n$. We have $$A(N) := \sum _{i \leq N}a_i = \sum_{n^k+m^k \leq N} n^{2k}$$ and similarly for $B$. We can estimate the sum from below and above by a Riemann integral over domains of the form $$D_{k, N} = \{(x,y) = 0 \leq x,y \;,\; x^k + y^k \leq N \}$$
Here are some nice pictures of what they look like: https://mathsci2.appstate.edu/~cookwj/maple/Circles/
It's not hard to see that
$$\begin{align*} A(N) &= \int_{D_{k,N}} x^{2k} dxdy + O_k(N^{2+1/k}) \\
&= \int_{0}^{N^{1/k}} x^{2k} (N-x^k)^{1/k}dx  + O_k(N^{2+1/k}) \\
&= N^{2+2/k} \int_0^1 u^{2k} (1-u^k)^{1/k}du  + O_k(N^{2+1/k}) \end{align*}$$
and similarly
$$B(N) = N^{2+2/k} \frac1{k+1}\int_0^1 u^k (1-u^k)^{1+1/k}du  + O_k(N^{2+1/k}) $$
We conclude that $$\begin{align*}
\lim_{N \to \infty}\frac{A(N)}{B(N)} &= (k+1)\frac{\int_0^1 u^{2k} (1-u^k)^{1/k}du}{\int_0^1 u^k (1-u^k)^{1+1/k}du} \\
&= (k+1) \frac{B\left(\frac{2k+1}k,1+ \frac{1}k\right)}{B\left(\frac{k+1}k, 2+\frac{1}k\right)} \\
&= k+1
\end{align*}$$
where $B$ is the Beta function, which satisfies $\int_0^1 u^a(1-u^k)^bdu = B(\frac{a+1}k, b+1)$ for $a,b > 0$.
Finally, $$p_k = \lim_{c \to 1^-}\alpha'(c) = 1- \lim_{N \to \infty}\frac{A(N)}{B(N)} = -k$$
