algorithms – Prove that if $k$ was the $(i+1)$st key to be inserted into the hash table, then $E[probes(k)]=frac{1}{1-frac{i}{m}}$

Theorem: Inserting an element into an open-address hash table with load factor α requires at
most $1/(1 − α)$ probes on average, assuming uniform hashing.

By following unsuccessful search strategy, we find out that average number of probes is $E(probesge i) = frac{1}{1-alpha}$ where $alpha = frac{n}{m}$, $n$ is elements and $m$ is slots.

Problem: Can you please help me to prove that if $k$ was the $(i +1)$st key inserted into the hash table, the expected number of probes made in a search for k is at most $1/(1 − (i)/m) = m/(m − i)$? Should it be that $1/(1 − (i+1)/m) = m/(m − (i+1))$ please?