Summation to n of Power of k over k/Proof 1

Proof
The proof proceeds by induction over $n$.

For all $n \in \Z_{\ge 1}$, let $\map P n$ be the proposition:
 * $\ds \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \ds \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {x - 1}^k} k$

Basis for the Induction
$\map P 1$ is the case:

Thus $\map P 1$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.

So this is the induction hypothesis:
 * $\ds \sum_{k \mathop = 1}^r \dfrac {x^k} k = H_r + \ds \sum_{k \mathop = 1}^r \dbinom r k \dfrac {\paren {x - 1}^k} k$

from which it is to be shown that:
 * $\ds \sum_{k \mathop = 1}^{r + 1} \dfrac {x^k} k = H_{r + 1} + \ds \sum_{k \mathop = 1}^{r + 1} \dbinom {r + 1} k \dfrac {\paren {x - 1}^k} k$

Induction Step
This is the induction step:

So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{\ge 1}: \ds \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \ds \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {x - 1}^k} k$