Prove $n + H(1) + H(2) + H(3) + ... + H(n-1) = nH(n)$ by induction

The question below is an extension question from an exercise booklet and I am unable to complete a small part of this problem at the end. I seem to be stuck on simplifying the left hand side during the induction proof.

Question:

Let $H(n) = 1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{n}$. Use induction to prove for all positive integers $n > 0$, $n + H(1) + H(2) + ... + H(n-1) = nH(n)$.

Attempt:

During the induction process:

Part A:

When $n = 1$, $RHS = 1×H(1)$

= $1 × 1$

= $1$

= $LHS$

so the statement is true for $n = 1$

Part B:

Prove for $k$ (Using the fact that n = k)

$$ k + H(1) + H(2) + ... + H(k - 1) = kH(k) $$

Prove for $k + 1$

$$ (k + 1) + H(1) + H(2) + ... + H(k) = (k + 1)H(k + 1) $$

Left hand side:

$ (k + 1) + H(1) + H(2) + ... + H(k)$

= $ kH(k) + H(k) + 1$ (** By induction hypothesis)

= $ H(k)(k+1) + 1 $

This is as far as I can progress. By substitution, it proves that, $ H(k)(k+1) + 1 $ = $ (k + 1)H(k+1)$, however, I am not sure how to simplify the left hand side further so it can equal the right hand side. I must be missing some important detail.

Can you please help?

$$ \begin{align} LHS &= (k+1)H(k) + 1 \\ &= (k+1)\left(1+\frac12+ \cdots+\frac1k\right) + 1 \\ &= (k+1)\left(1+\frac12+ \cdots+\frac1k\right) + (k+1)\left(\frac{1}{k+1}\right) \\ &= (k+1)\left(1+\frac12+ \cdots+\frac1k + \frac{1}{k+1}\right) \\ &= (k+1)H(k+1) \\ &= RHS \end{align} $$

Here is a direct proof that you may find interesting. Let $n\geq 2 $, we have : \begin{aligned}\sum_{k=1}^{n-1}{H_{k}}&=\sum_{k=1}^{n-1}{\left(\left(k+1\right)H_{k}-kH_{k}\right)}\\ &=\sum_{k=1}^{n-1}{\left(\left(k+1\right)\left(H_{k+1}-\frac{1}{k+1}\right)-kH_{k}\right)}\\ &=\sum_{k=1}^{n-1}{\left(\left(k+1\right)H_{k+1}-kH_{k}\right)}-\sum_{k=1}^{n-1}{1}\\ \sum_{k=1}^{n-1}{H_{k}}&=nH_{n}-n\end{aligned}