The proof is from Ross's A Second Course in Probability.
By definition, $A_{k,n}=q_nA_{k-1,n}+p_nA_{k-1,n-1}$ is $P(X_1+...+X_n>k)=P(X_n>1)P(X_1+...+X_n>k-1)+P(X_n=1)P(X_1+...+X_{n-1}>k-1)$
Conditioning on $X_n>1$ not occuring, $X_n$ must be $1$, then $X_1+...+X_n>k$ is equivalent to $X_1+...+X_{n-1}>k-1$. The second part is pretty straightforward.
What troubles me is the first summand. Assuming $X_n>1$, how to decude $X_1+...+X_n>k$ from $X_1+...+X_n>k-1$?
At the heart of this is the following identity which holds for a single geometric random variable $X$ with success probability $0 < p < 1$, equivalently failure $q = 1-p$,
Claim For $k > 1$, $$ \mathbf P[X > k | X > 1] = \mathbf P[X > k-1].$$
You can prove this fairly immediately by appealing to Bayes theorem, and the formula for the CDF of a Geometric random variable:
$$ \begin{aligned}\mathbf P[X > k | X > 1] & = \mathbf P[X > 1 | X > k ] \frac{\mathbf P[X > k]}{\mathbf P[X > 1]} \\ & = 1 \times \frac{q^k}{q} \\ & = q^{k-1} \\ & = \mathbf P[X > k-1]. \end{aligned}$$
In fact the identity has an intuitive explanation, detailed at the end.
If you follow Ross's conditioning argument for the case that $n = 1$, you will see that this claim is all that is required to derive the identity.
Extending to the case n > 1
The identity which you actually are trying to prove is the more general statement
$$ \mathbf P[X_1 + \cdots + X_n > k|X_n > 1] = \mathbf P[X_1 + \cdots + X_n > k-1]$$
The key here is to recognise that the sum $S_{n-1} = X_1 + \cdots + X_{n-1}$ is independent of $X_n$ and therefore just represents noise. Eg. if we assume this term is a constant $S_{n-1} = s$ then the identity says
$$ \mathbf P[s + X_n > k | X_n > 1] = \mathbf P[s + X_n > k-1],$$
and moving the term in $s$ to the other side, and considering $k' = k-s$ then we're done!
All that is left is to make this argument water tight; there are two issues to handle:
We cannot treat $S_{n-1} = s$ as a constant quite this easily and we need to formalise the independence argument.
We need to be careful to take account of cases depending on whether $s > k$ or not.
The second point is just being careful about ranges that you sum over, so is easily resolved, whilst the first point follows by using independence and applying the formula that follows from the law of total probability.
Hopefully that's enough for you to fill in the detail yourself.
Intuition for Claim
If we interpret $X$ as the number of trials before we achieve a heads when throwing a (biased) coin, and if we rewrite the left hand side of the above as the equivalent identity $$ \mathbf P[X > k | X > 1] = \mathbf P[X > k | X \neq 1]$$
Then this says, what's the probability we must throw the coins $k$ times in total given we didn't get a head on the first throw.
This is the same as assuming the first throw never happened, and asking what's the probability it will take $k-1$ throws, which is exactly the right hand side.