$f(n+1) = f(n) + f(n)^{a}$ where $a \in (0,1)$ and $n \ge 1$ with $f(1) = m$.
If $a=0$, we see $f(n) = m + n - 1$ and if $a=1$, we see $f(n) = 2^{n-1}m$. So the recursion seems to interpolate between linear and exponential forms.
Is there a closed form or tight approximation or tight asymptotics for $f(n)$ in terms of $n$, $a$ and $m$?
Update after leshik's answer
Can the same asymptotics hold if $a \ge 1$? What is the asymptotics/tight approximation/exact expression if $a \ge 1$?
We "guess" the asymptotic by making use of Stolz-Cesaro theorem. The idea is to show that the sequence behaves like $n^{\alpha}$ for an appropriate choice of $\alpha.$ By Stolz- Cesaro, $$A=\lim_{n\to\infty}\frac{x_{n+1}}{(n+1)^{\alpha}}=\lim_{n\to\infty}\frac{x_{n+1}-x_n}{(n+1)^{\alpha}-n^{\alpha}}=\lim_{n\to\infty}\frac{x_n^a}{\alpha n^{\alpha-1}}=\frac{1}{\alpha}\lim_{n\to\infty}\left(\frac{x_n}{ n^{{\alpha-1}/a}}\right)^a.$$
Now, we want to have $\frac{\alpha-1}{a}=\alpha$ or $\alpha=\frac{1}{1-a}.$ The equation for our limit can be rewritten as $A=\frac{1}{\alpha}A^a$ which gives $A=(1-a)^{\frac{1}{1-a}}.$