Let be a sequence $\{ y_k \}^\infty _{k=0} \subset (0,\infty) $ that satisfies $$ y_k\le {2^k\over M}y_{k-1}^\beta , $$ where $k=1,2,...$, and $\beta\gt 1$ , $M\gt0$.
Prove that if $M\gt2^{\beta\over \beta-1}y_0^{\beta-1}$, then $\lim_{k\to \infty}y_k=0$.
Try to compare with $$ ca^ky_k=u_k\le u_{k-1}^β= (ca^{k-1}y_{k-1})^β \\\iff\\ y_k\le c^{β-1}a^{(k-1)β-k}y_{k-1}^β=(c^{β-1}a^{-β})(a^{β-1})^ky_{k-1}^β $$ which is successful by identifying the relations $2=a^{β-1}$, $M^{-1}= a^{-β}c^{β-1}$, that is $$ a=2^{\frac1{β-1}},\quad c=M^{-\frac1{β-1}}2^{\frac{β}{(β-1)^2}} $$ It follows that $$ u_n\le u_0^{β^n} $$ which converges to zero for $0\le u_0<1$, i.e., $$ 0\le M^{-\frac1{β-1}}2^{\frac{β}{(β-1)^2}}y_0<1 $$ which can be transformed into the given condition.
Since $a>1$ the sequence $y_k$ converges to zero even if $u_k$ is bounded, so that also equality in the condition is permissible.