Questions on the proof of make change problem using greedy algorithm I recently learned the proof of the greedy algorithm solution to make change problem:

There's a set of coins $\{c_1 < c_2<...<c_{i-1}<c_i\}$. $c_i$ is an integer and $c_{i+1}=q\cdot c_i$ where $q$ is an integer. Given number $x$ what is the least amount of coins needed to pay $x$ of some currency?

The solution will use a greedy algorithm. In order to prove that greedy algorithm is correct we suppose that we have:
$$
\{c_1, c_2,...,c_{i-1},c_i\} \quad\text{coins set}\\
k_1, k_2,...,k_{i-1},k_i\quad\text{how many times}\quad c_i \quad\text{is repeated}\\
c_1=1
$$
We also suppose that there's another different algorithm that works and in it there're $p_i$
repetitions of $c_i$. 
Then $p_i<k_i$. Then we have some "deficit" on $p_i$ part then the deficit must be resolved by some former combination of $p$ and $c$. 
We'll prove that the deficit will never be resolved because:
$$
p_1\cdot c_1<c_2\qquad (1)
$$
then:
$$
p_1\cdot c_1 \le c_2-c_1 \implies\qquad (2)\\
p_2\cdot c_2 \le c_3-c_2 \implies\\
\vdots\\
p_i\cdot c_i \le c_i-c_{i-1} \implies
$$
Now if we sum up the LHS and RHS we get:
$$
\sum_{k=1}^{i-1}p_k\cdot c_k\le c_i-c_1<c_i
$$
which proves that the deficit is never made up for by $\sum p_k\cdot c_k$.

Two things I can't grasp:
why does $(1)$ and $(2)$ hold?

(1) is true because you could have used $c_2$'s instead if $p_1c_1 \geq c_2$.
(2) is true from (1) because $qc_1 = c_2$ for some integer $q$, so if $p_1c_1 < c_2$ then $(p_1 + 1)c_1 \leq c_2$
