Erlang induction step

Derive the formula:

“Exp(λ)+Erlang(ℓ,λ)∼Erlang(ℓ+1,λ)”

Observation: By repeatedly applying the above formula, we see that:

“Exp(λ)+⋯+Exp(λ)⏞ℓ terms∼Erlang(ℓ,λ)”