Transform grammar for repeating characters into LL(1)
Let's construct an LL(1) grammar for the language $\mathscr{L}(a^+)$.

Firstly, observe that all the production rules in this case must have one of two forms:
$$A \to a\Gamma$$
or
$$A \to \epsilon,$$
where $\Gamma$ is a (possibly empty) sequence of non-terminals and the terminal 'a'.

The first step would be adding the rule $S \to aS'$, since it's clear that $S \to aS \mid a$ doesn't give us an LL(1) grammar.
Obviously $S'$ must have a production rule with the empty string as its body $S' \to \epsilon$, which will be used when we get the end of string symbol $\$$ during parsing. We also need a "looping construct" to finish our construction, resulting in the following LL(1) grammar:

$$\begin{align}
S  &amp;\to aS' \\
S' &amp;\to \epsilon \mid aS'
\end{align}$$

From wikipedia:


  A nonterminal is left-recursive if the leftmost symbol in one of its productions is itself (...) or can be made itself by some sequence of substitutions (...).


Since neither $S'$ nor $S$'s left-most non-terminal symbol cannot be 'made itself' by any sequence of substitutions (there are simply no left-most non-terminal symbols there), the resulting grammar is not left-recursive.