Inequality proof concerning two variables $k,q$ Let $q,k$ positive integers satisfying the following inequalities
$$k\geq3$$
$$q\geq2$$
$$k\geq q$$
I have checked that the following inequality holds for a few values of $k,q$ I have tried but I have not been able to prove it.
$$A\triangleq kq{{q^{k-2}}\choose{q^{k-3}+1}}(k-1)\geq kq\cdot q^{2k-4}(q-1)-q^{k-1}(q-1)\frac{(q-1)^3+1}{q}k\triangleq B$$
I tried to simplify $B$ as
\begin{eqnarray*}
B&=&kq\cdot q^{2k-4}(q-1)-(q^k-q^{k-1})\frac{q^3-3q^2+3q}{q}k\\
&=&kq\cdot q^{2k-4}(q-1)-(q^k-q^{k-1})(q^2-3q+3)k\\
&=&kq\cdot q^{2k-4}(q-1)-(q^{k+2}-3q^{k+1}+3q^k-q^{k+1}+3q^k-3q^{k-1})k\\
&=&kq\cdot q^{2k-4}(q-1)-kq^{k+2}+4kq^{k+1}-6kq^k+3kq^{k-1}\\
&=&kq\cdot(q^{2k-3}-q^{2k-4}-q^{k+1}+4q^k-6q^{k-1}+3q^{k-2})\\
\end{eqnarray*}
Now, I need to show that 
$${{q^{k-2}}\choose{q^{k-3}+1}}(k-1)\geq(q^{2k-3}-q^{2k-4}-q^{k+1}+4q^k-6q^{k-1}+3q^{k-2})$$
and I am stuck at this point.
 A: The inequality does hold for $k\ge 3$, $q\ge 2$ and $k\ge q$.

We can directly see that the inequality holds for $(k,q)=(3,2),(3,3),(4,2),(4,3),(4,4),(5,2),(6,2)$ respectively.
Here, we use two lemmas.
Lemma 1 : ${{q^{k-2}}\choose{q^{k-3}+1}}(k-1)\ge q^{q^{k-3}}(q-1)$.
Proof for lemma 1 :
$$\begin{align}&{{q^{k-2}}\choose{q^{k-3}+1}}(k-1)
\\\\&\ge {{q^{k-2}}\choose{q^{k-3}+1}}(q-1)
\\\\&=\frac{q^{k-2}\times (q^{k-2}-1)\times \cdots\times (q^{k-2}-q^{k-3})}{(q^{k-3}+1)\times  q^{k-3}\times \cdots \times 1}\times(q-1)
\\\\&=\frac{q^{k-2}}{q^{k-3}+1}\times \frac{q^{k-2}-1}{q^{k-3}}\times \frac{q^{k-2}-2}{q^{k-3}-1}\times\frac{q^{k-2}-3}{q^{k-3}-2}\times \cdots\times\frac{q^{k-2}-q^{k-3}}{1}\times (q-1)
\\\\&\ge\frac{q^{k-2}}{q^{k-3}+1}\times \frac{q^{k-2}-1}{q^{k-3}}\times q\times q\times\cdots\times q\times q\times (q-1)
\\\\&=\frac{q^{k-2}}{q^{k-3}}\times\frac{q^{k-2}-1}{q^{k-3}+1}\times q^{q^{k-3}-1}\times (q-1)
\\\\&\ge q\times 1\times q^{q^{k-3}-1}\times (q-1)
\\\\&=q^{q^{k-3}}(q-1)\quad\square
\end{align}$$
Lemma 2 : $8\cdot q^{2k-3}\geq q^{2k-3}-q^{2k-4}-q^{k+1}+4q^k-6q^{k-1}+3q^{k-2}$.
Proof for lemma 2 : 
$$\begin{align}8\cdot q^{2k-3}&=q^{2k-3}+4q^{2k-3}+3q^{2k-3}
\\\\&\ge q^{2k-3}+4q^{k}+3q^{k-2}
\\\\&\ge q^{2k-3}-q^{2k-4}-q^{k+1}+4q^k-6q^{k-1}+3q^{k-2}\quad\square
\end{align}$$
Now, from the two lemmas, in order to prove the inequality, it is sufficient to prove
$$q^{q^{k-3}}(q-1)\ge 8\cdot q^{2k-3}\tag1$$
We can directly see that $(1)$ holds for $(k,q)=(5,3),(5,4),(5,5),(6,3),(6,4),(6,5),(6,6)$ respectively.
Lemma 3 : $(k-3)\log 2-\log(2k)\ge 0$ for $k\ge 7$.
Proof for lemma 3 : Let $f(k)=(k-3)\log 2-\log(2k)$. Then, $f'(k)=\log 2-\frac 1k$. So, $f(k)$ is increasing for $k\ge 7$ with $f(7)=\log\frac 87\gt 0$. $\ \ \square$ 
So, for $k\ge 7$, $q\ge 2$ and $k\ge q$, we can get, from the lemma 3, 
$$\begin{align}\log q\ge \log 2\ge\frac{\log(2k)}{k-3}
&\implies(k-3)\log q\ge\log(2k)
\\\\&\implies q^{k-3}\ge 2k
\\\\&\implies q^{k-3}-2k+3\ge 3
\\\\&\implies q^{q^{k-3}-2k+3}\ge q^3\ge 8
\\\\&\implies q^{q^{k-3}-2k+3}(q-1)\ge 8(q-1)\ge 8
\\\\&\implies q^{q^{k-3}}(q-1)\ge 8\cdot q^{2k-3}\end{align}$$
which is $(1)$.
Therefore, the inequality holds for $k\ge 3$, $q\ge 2$ and $k\ge q$.
