$f(n+1) = f(n) + f(n)^{a}$ 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 for $f(n)$ in terms of $n$, $a$ and $m$?