INTRODUCTION
11, 111, 1111 in base ten are not pure powers, we can easily verify it, but
is it true for a number of n digits equal 1 in base ten, for all n. This study
is an answer to this question. A number with n digits equal to 1 (111…1)
in base ten is equal to a number with n digits equal to 9 (999…9) in base
ten divided by 9, which means ,
the theorem stipulates that ,
.
This research proposes an original and analytic proof of this theorem called
Bugeaud and Mignotte theorem (Bugeaud and Mignotte, 1999).
THE PROOF
Let BugeaudMignotte equation:

Then with j^{2} = 1 
10^{n}9Y^{q} = 1 = 10^{n}+j(9jY^{q}) 
We pose: 
10^{n} = x, 9jY^{q} = y, u = 1, z = 9jY^{q}10^{n} 
Lemma 1
Proof of Lemma 1

And: 

We pose: 

until infinty. 
Lemma 2
The expression of the sequences is 

Proof of Lemma 2
For I = 2, ,
it is verified. We suppose the expressions true for I 

But: 

Lemma 3
x_{i}jy_{i} = xjy
Lemma 4
The lemmas 1, 2, 3 imply that jyx = 0
Proof of Lemma 4

We pose: 

We pose: 


And: 

And: 

If we make the hypothesis that 

And: 


But: 

The hypothesis
is false and 

In all cases 


is convergent, then the general term of the series tends to zero 

And it is impossible: the existence of Y and q is impossible! 
CONCLUSION
BugeaudMignotte equation has effectively no solution, and an analytic proof
exists. The generalization is
is it possible? It seems that there are only three solutions: