Lean/Mathlib: prove ∑ aₙsⁿ = exp(∑ cₖsᵏ/k) from the recursion (n+1)aₙ₊₁ = ∑ cₖ₊₁aₙ₋ₖ by an ODE argument

Mathlib has no exp of a power series as a function identity, but the calculus route is short. Shift indices so every series is ∑ bₙ z^(n+1): then hasDerivAt_tsum_of_isPreconnected on t = Ioo (-r) r with r = (|y|+1)/2, bound u n = (n+1) rⁿ (summable_pow_mul_geometric_of_norm_lt_one 1 plus geometric), gives the termwise derivative; pass g and g' explicitly as named args or the summability side goal at y₀ = 0 is a metavariable and simp fails. For A = ∑ aₙzⁿ use Summable.tsum_eq_zero_add and HasDerivAt.congr_of_eventuallyEq. The recursion is exactly A' = C'A via tsum_mul_tsum_eq_tsum_sum_range_of_summable_norm, so (A·exp(-C))' = 0 and IsOpen.is_const_of_deriv_eq_zero isOpen_Ioo isPreconnected_Ioo makes it constant = 1. To then let s → 1⁻, Real.tendsto_tsum_powerSeries_nhdsWithin_lt (Abel) applied to the differences u n - u (n-1) gives (1-s)∑uₙsⁿ → lim u.