Prove that if you can derive w from α in n steps, it's possible with n left-derivations as well
Your method of proof can't work, since it implies that every derivation is a leftmost derivation. Indeed, here is a paraphrase of your proof:


  Let's prove by induction on the length of the proof that if a word $w$ can be derived from a sentential form $\alpha$ in $n$ steps, then $w$ can be derived from $\alpha$ in $n$ steps using a leftmost derivation.
  The case $n = 0$ is trivial.
  Now suppose that $n &gt; 0$, and suppose that the first step of the derivation is $\alpha \Rightarrow \beta$. By induction, $\beta \Rightarrow^* w$ can be converted to a leftmost derivation of length $n-1$, so $\alpha \Rightarrow \beta \Rightarrow^* w$ would be a leftmost derivation of length $n$.


Unrolling the proof, your suggested derivation is actually the same as the original derivation (prove it by induction!), and since not all derivations are leftmost, since can't work.

Concretely, consider the derivation $AB \Rightarrow Ab \Rightarrow ab$. The first step $AB \Rightarrow Ab$ cannot belong to a leftmost derivation. Answering your question, you can't prove that $\alpha \stackrel{lm^1}{\Rightarrow} \beta$, simply because it isn't true.



Here is a different approach. Consider the derivation tree used to derive $\alpha$ from $w$. Try to use this tree to generate a leftmost derivation. Note that the length of the derivation is exactly the number of nonleaf nodes in the tree.