Proof.
Let
g_{r} g_{r1} 
(
g_{r1})
s_{r1}mod
p^{2r}.
Then we have
g g_{r} modp^{} 
(124) 
In other words
g_{r} is a
solution of higher precision.
Then proving the theorem turns into proving the invariants for
i = 0
^{ ... }r

g_{i} g_{0}modp,

(g_{i}) 0 modp^{2i},
 if i < r then
s_{i} ^{}(g_{i}) 1 modp^{2i}.
The case
i = 0 is clear and we assume that
i > 0 holds.
Then by induction hypothesis
p^{2i1} divides

(g_{i1})

s_{i1}  ^{}(g_{i1})^{1}
Then
p^{2i} divides their product, that is
(g_{i1}s_{i1} (g_{i1})^{}(g_{i1})^{1} modp^{2i}. 
(125) 
Now we calculate
g_{i} 

g_{i1}  (g_{i1})s_{i1} modp^{2i} 


g_{i1}  (g_{i1})^{}(g_{i1})^{1} modp^{2i} 

(126) 
Now applying Proposition
10
with
g =
g_{i},
h =
g_{i} and
k = 2
^{i1}
we obtain the first two invariants.
...