Show $|u^n| = n|u|$ for all strings $u$ and all $n$ Can anyone please help me with this homework question on automata from Peter Linz?
Use induction on $n$ to show that $|u^n| = n|u|$ for all strings $u$ and all $n$.
 A: First we need to know what the cardinality of a string is.  For example what is the cardinality of the string 'xyz'.  I would guess it's 3 in this case (the number of characters in the string), but this should be confirmed.
Next we need to know how $u^n$ is defined. Again, it might be helpful to consider the example 'xyz'. I would guess $(xyz)^2$ would be xyzxyz, so $u^n$ is n lots of the string u concatenated together into one string.
Based on these assumptions, the proof by induction would be constructed in this way:
Base case $(n=1):$ $|u^1| = |u| = 1|u|$
Inductive step: 
Suppose $|u^k|=k|u|$
Then $|u^{k+1}| = |u^k| + |u|$ (by definitions of cardinality of string and $u^n$)
= k|u| +|u| (employing the inductive hypothesis)  
= (k+1)|u| (by definition of cardinality of string)  
The result follows by induction
I'm not sure if $n = 0$ needs to be considered. If there is a definiton of $u^0$ it woild presumably be  $u^0 =$ the empty string $(\lambda)$, the cardinality of which I would presume to be $0$, so the base case would then be
Base case $(n=0):$ $|u^0| = |\lambda| = 0 = 0|u|$ so it would still work and the rest of the proof would be identical.
