Hint:
Consider the slightly stronger statement:
$A(m,n+1)\ge A(m,n)+2$ for all $m,n\in\mathbb N$.
Note that this also implies $A(m,n)\ge2n$ by induction over $n$.
Prove it by induction over $m$.
Base case $m=0$ follows trivially.
Inductive case: Suppose it holds for some $m=j$. We prove it holds for $m=j+1$. We prove by induction over $n$.
Base case $n=0$ follows trivially.
Base case $n=1$ follows trivially. (Show that $A(m,2)=4$ for all $m$.)
Inductive case: Suppose it holds for some $n=k\ge1$. We prove it holds for $n=k+1$::
$A(j+1,k+2)\\=A(j,A(j+1,k+1))\\\ge2A(j+1,k+1)\\=A(j+1,k+1)+A(j+1,k+1)\\\ge A(j+1,k+1)+2(k+1)\\\ge A(j+1,k+1)+2.$
I will show for 2 indices instead of 9, as it is simpler to see. But firstly I want to say that in Mathematics often symbols are very very ugly, but once you decompose what each part means it's quite simple.
$$\displaystyle\bigwedge\limits_{i=1}^2\bigwedge\limits_{n=1}^2\bigvee\limits_{j=1}^2p(i,j,n)$$
Let us decompose this! Maybe in this form, it will already be obvious what this means.
$$
\bigwedge\limits_{i=1}^2\left(\bigwedge\limits_{n=1}^2\left( \bigvee\limits_{j=1}^2p(i,j,n) \right)\right)
$$
If not, then let us decompose it, first we evaluate the inner parentheses:
$$
\bigwedge\limits_{i=1}^2\left(\bigwedge\limits_{n=1}^2[ p(i,1,n) \vee p(i,2,n) ]\right)
$$
Then we evaluate the second parentheses:
$$
\bigwedge\limits_{i=1}^2 \Big( [ p(i,1,1) \vee p(i,2,1) ] \wedge [ p(i,1,2) \vee p(i,2,2) ] \Big)
$$
Then finally we examine the last statement:
$$
\Big( [ p(1,1,1) \vee p(1,2,1) ] \wedge [ p(1,1,2) \vee p(1,2,2) ] \Big) \wedge \Big([ p(2,1,1) \vee p(2,2,1) ] \wedge [ p(2,1,2) \vee p(2,2,2) ] \Big)
$$
If these expansions confuse you, a trick I used when I started with math several years ago is just imagine you have all the terms specified by the subscripts and superscripts, then 'combine' them with the operation given (the large symbol on the left, for example, $\Sigma$ means after having all the terms, we add all of them). Of course there are (not so) minor details like commutativity/ability to re-arrange terms or not etc.
Best Answer
We have:
\begin{align} [2-2*7+2*7^2-...+2*(-7)^{k}]+2*(-7)^{k+1} &=\frac{1-(-7)^{k+1}+8(-7)^{k+1}}{4}\\ &=\frac{1+8(-7)^{k+1}-1(-7)^{k+1}}{4} \end{align} Can you take it from here?