Prove that exponent is less than number of digits of a perfect power?

Question

I am trying to find a proof for below statement. i.e. Given integer N that is a perfect power, show that the exponent is less than the number of digits in N

Show that if N = M^e for some integers M,e > 1 then e ≤ |N| + 1

Not able to find any hints for the same. Any help?

FYI. This question if from book : 'Introduction to Modern Cryptography, Second Edition By Jonathan Katz, Yehuda Lindell' Exercise 8.12 a)

https://books.google.co.in/books?id=OWZYBQAAQBAJ&pg=PA339&lpg=PA339&dq=THis+exercise+develops+an+efficient+algorithm+for+testing&source=bl&ots=BbpgCOX_2m&sig=B8y4vfYjES-hqDV_nvup0gZa2yg&hl=en&sa=X&ved=0ahUKEwiS6ODT4L7LAhWKC44KHb3LCfcQ6AEIGzAA#v=onepage&q=THis%20exercise%20develops%20an%20efficient%20algorithm%20for%20testing&f=false

Answer 1

If we let $||N||$ represent the number of digits in $N$, then we have:$$\begin{align} N&=M^e\\ \therefore \log_{10}N&=e\log_{10}M\\ \therefore e&=\frac{\log_{10}N}{\log_{10}M}\lt\frac{||N||}{\log_{10}M}\\ \end{align}$$Now this will give $e\lt||N||$ only when $\log_{10}M\ge1$ which means it is only valid for $M\ge10$

Answer 2

If you take $N = 16$, $M = 2$, and $e = 4$, we have $N = 2^4$ and $N$ has two digits. We see that $e > 2$, and the claim is false.

Answer 3

Recall that a number $A$ has $d$ digits if and only if $$2^{d-1} \le A < 2^d-1 $$ Now, if $M \ge 2$ you have $$M^e \ge 2^e$$ so that $M^e$ as at least $e+1$ digits.