We compare $\lfloor a^k\rfloor$ with $\lfloor a\rfloor^k$ in $3$ possible cases: Case 1) $0<a<1\Longrightarrow 0<a^k<1, \forall k\in\mathbb{N}\Longrightarrow \lfloor a^k\rfloor=\lfloor a\rfloor^k=0, \forall k\in\mathbb{N}$. Case 2) $a\in\mathbb{N}\Longrightarrow \lfloor a^k\rfloor=\lfloor a\rfloor^k=a^k, \forall k\in\mathbb{N}$. Case 3) $a=m+\alpha$, where $m\in\mathbb{N}, 0<\alpha<1$. $a^k=(m+\alpha)^k\ge m^k+km^{k-1}\alpha\ge m^k+k\alpha$. For $k\ge \dfrac{1}{\alpha}+1$ results: $a^k>m^k+1\Longrightarrow\lfloor a^k\rfloor\ge \lfloor m^k+1\rfloor=\lfloor a\rfloor^k+1>\lfloor a\rfloor^k$, hence the equality $\lfloor a^k\rfloor=\lfloor a\rfloor^k$ occurs for maximum $\dfrac{1}{\alpha}$ values of $k$. If the equality $\lfloor a^k\rfloor+\lfloor b^k\rfloor=\lfloor a\rfloor^k+\lfloor b\rfloor^k$ occurs for an infinite number of $k$, results: $a\in(0,1)\cup\mathbb{N};b\in(0,1)\cup\mathbb{N}\Longrightarrow$ $\Longrightarrow \lfloor a^k\rfloor=\lfloor a\rfloor^k$ and $\lfloor b^k\rfloor=\lfloor b\rfloor^k, \forall k\in\mathbb{N}\text{ (inclusive for k=2014) }\Longrightarrow$ $\Longrightarrow \lfloor a^{2014}\rfloor+\lfloor b^{2014}\rfloor=\lfloor a\rfloor^{2014}+\lfloor b\rfloor^{2014}$.