/-- Alternate form of **Faulhaber's theorem**, relating the sum of p-th powers to the Bernoulli
numbers: $$\sum_{k=1}^{n} k^p = \sum_{i=0}^p (-1)^iB_i\binom{p+1}{i}\frac{n^{p+1-i}}{p+1}.$$
Deduced from `sum_range_pow`. -/
theorem sum_Ico_pow (n p : ℕ) :
    (∑ k in Ico 1 (n + 1), (k : ℚ) ^ p) =
      ∑ i in range (p + 1), bernoulli' i * (p + 1).choose i * (n : ℚ) ^ (p + 1 - i) / (p + 1) := by
  rw [← Nat.cast_succ]
  -- dispose of the trivial case
  cases' p with p p
  · simp
  let f i := bernoulli i * p.succ.succ.choose i * (n : ℚ) ^ (p.succ.succ - i) / p.succ.succ
  let f' i := bernoulli' i * p.succ.succ.choose i * (n : ℚ) ^ (p.succ.succ - i) / p.succ.succ
  suffices (∑ k in Ico 1 n.succ, (k : ℚ) ^ p.succ) = ∑ i in range p.succ.succ, f' i by convert this
  -- prove some algebraic facts that will make things easier for us later on
  have hle := Nat.le_add_left 1 n
  have hne : (p + 1 + 1 : ℚ) ≠ 0 := by norm_cast; exact succ_ne_zero p.succ
  have h1 : ∀ r : ℚ, r * (p + 1 + 1) * (n : ℚ) ^ p.succ / (p + 1 + 1 : ℚ) = r * (n : ℚ) ^ p.succ :=
      fun r => by rw [mul_div_right_comm, mul_div_cancel _ hne]
  have h2 : f 1 + (n : ℚ) ^ p.succ = 1 / 2 * (n : ℚ) ^ p.succ := by
    simp_rw [bernoulli_one, choose_one_right, succ_sub_succ_eq_sub, cast_succ, tsub_zero, h1]
    ring
  have :
    (∑ i in range p, bernoulli (i + 2) * (p + 2).choose (i + 2) * (n : ℚ) ^ (p - i) / ↑(p + 2)) =
      ∑ i in range p, bernoulli' (i + 2) * (p + 2).choose (i + 2) * (n : ℚ) ^ (p - i) / ↑(p + 2) :=
    sum_congr rfl fun i _ => by rw [bernoulli_eq_bernoulli'_of_ne_one (succ_succ_ne_one i)]
  calc
    (-- replace sum over `Ico` with sum over `range` and simplify
        ∑ k in Ico 1 n.succ, (k : ℚ) ^ p.succ) = ∑ k in range n.succ, (k : ℚ) ^ p.succ :=
      by simp [sum_Ico_eq_sub _ hle, succ_ne_zero]
    -- extract the last term of the sum
        _ = (∑ k in range n, (k : ℚ) ^ p.succ) + (n : ℚ) ^ p.succ :=
      by rw [sum_range_succ]
    -- apply the key lemma, `sum_range_pow`
        _ = (∑ i in range p.succ.succ, f i) + (n : ℚ) ^ p.succ :=
      by simp [sum_range_pow]
    -- extract the first two terms of the sum
        _ = (∑ i in range p, f i.succ.succ) + f 1 + f 0 + (n : ℚ) ^ p.succ :=
      by simp_rw [sum_range_succ']
        _ = (∑ i in range p, f i.succ.succ) + (f 1 + (n : ℚ) ^ p.succ) + f 0 := by ring
        _ = (∑ i in range p, f i.succ.succ) + 1 / 2 * (n : ℚ) ^ p.succ + f 0 := by 

Proof state:
A : Type u_1
inst✝¹ : CommRing A
inst✝ : Algebra ℚ A
n p : ℕ
f : ℕ → ℚ := fun i => bernoulli i * ↑(Nat.choose (succ (succ p)) i) * ↑n ^ (succ (succ p) - i) / ↑(succ (succ p))
f' : ℕ → ℚ := fun i => bernoulli' i * ↑(Nat.choose (succ (succ p)) i) * ↑n ^ (succ (succ p) - i) / ↑(succ (succ p))
hle : 1 ≤ n + 1
hne : ↑p + 1 + 1 ≠ 0
h1 : ∀ (r : ℚ), r * (↑p + 1 + 1) * ↑n ^ succ p / (↑p + 1 + 1) = r * ↑n ^ succ p
h2 : f 1 + ↑n ^ succ p = 1 / 2 * ↑n ^ succ p
this :
  ∑ i in range p, bernoulli (i + 2) * ↑(Nat.choose (p + 2) (i + 2)) * ↑n ^ (p - i) / ↑(p + 2) =
    ∑ i in range p, bernoulli' (i + 2) * ↑(Nat.choose (p + 2) (i + 2)) * ↑n ^ (p - i) / ↑(p + 2)
⊢ ∑ i in range p, f (succ (succ i)) + (f 1 + ↑n ^ succ p) + f 0 =
    ∑ i in range p, f (succ (succ i)) + 1 / 2 * ↑n ^ succ p + f 0

Next tactic:
rw [h2]