Theorem: Summation to n of Power of k over k

:$\ds \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {x - 1}^k} k$
where:
:$H_n$ denotes the $n$th harmonic number
:$\dbinom n k$ denotes a binomial coefficient.

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:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^1 \dfrac {x^k} k
      | r = \dfrac {x^1} 1
      | c = 
}}
{{eqn | r = x
      | c = 
}}
{{eqn | r = 1 + \paren {x - 1}
      | c = 
}}
{{eqn | r = 1 + \dbinom 1 1 \paren {x - 1}
      | c = Binomial Coefficient with Self, for example
}}
{{eqn | r = H_1 + \sum_{k \mathop = 1}^1 \dbinom 1 k \dfrac {\paren {x - 1}^k} k
      | c = Harmonic Number H1
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \sum_{k \mathop = 1}^{r + 1} \dfrac {x^k} k
      | r = \sum_{k \mathop = 1}^r \dfrac {x^k} k + \dfrac {x^{r + 1} } {r + 1}
      | c = 
}}
{{eqn | r = H_r + \sum_{k \mathop = 1}^r \dbinom r k \dfrac {\paren {x - 1}^k} k + \dfrac {x^{r + 1} } {r + 1}
      | c = Induction Hypothesis
}}
{{eqn | r = H_r + \sum_{k \mathop = 1}^r \dbinom r k \dfrac {\paren {x - 1}^k} k + \dfrac {\paren {x^{r + 1} - 1} + 1} {r + 1}
      | c = 
}}
{{eqn | r = H_r + \sum_{k \mathop = 1}^r \dbinom r k \dfrac {\paren {x - 1}^k} k + \dfrac {\paren {x^{r + 1} - 1} } {r + 1} + \dfrac 1 {r + 1}
      | c = 
}}
{{eqn | r = H_{r + 1} + \sum_{k \mathop = 1}^r \dbinom r k \dfrac {\paren {x - 1}^k} k + \dfrac {x^{r + 1} - 1} {r + 1}
      | c = 
}}
{{eqn | r = H_{r + 1} + \sum_{k \mathop = 1}^r \paren { \dbinom {r + 1} k - \dbinom r {k - 1} } \dfrac {\paren {x - 1}^k} k + \dfrac {x^{r + 1} - 1} {r + 1}
      | c = Pascal's Rule
}}
{{eqn | r = H_{r + 1} + \sum_{k \mathop = 1}^r \dbinom {r + 1} k \dfrac {\paren {x - 1}^k} k - \sum_{k \mathop = 1}^r \dbinom r {k - 1} \dfrac {\paren {x - 1}^k} k + \dfrac {x^{r + 1} - 1} {r + 1}
      | c =
}}
{{eqn | r = H_{r + 1} + \sum_{k \mathop = 1}^r \dbinom {r + 1} k \dfrac {\paren {x - 1}^k} k - \sum_{k \mathop = 1}^r \dfrac k {r + 1} \dbinom {r + 1} k \dfrac {\paren {x - 1}^k} k + \dfrac {x^{r + 1} - 1} {r + 1}
      | c = Factors of Binomial Coefficient
}}
{{eqn | r = H_{r + 1} + \sum_{k \mathop = 1}^r \dbinom {r + 1} k \dfrac {\paren {x - 1}^k} k - \dfrac 1 {r + 1} \sum_{k \mathop = 1}^r \dbinom {r + 1} k \paren {x - 1}^k + \dfrac {x^{r + 1} - 1} {r + 1}
      | c =
}}
{{eqn | r = H_{r + 1} + \sum_{k \mathop = 1}^r \dbinom {r + 1} k \dfrac {\paren {x - 1}^k} k - \dfrac 1 {r + 1} \paren {\paren {\paren {x - 1} + 1 }^{r + 1} - 1 - \paren {x - 1}^{r + 1} } + \dfrac {x^{r + 1} - 1} {r + 1}
      | c = Binomial Theorem
}}
{{eqn | r = H_{r + 1} + \sum_{k \mathop = 1}^r \dbinom {r + 1} k \dfrac {\paren {x - 1}^k} k - \dfrac {x^{r + 1} - 1} {r + 1} + \dfrac {\paren {x - 1}^{r + 1} } {r + 1} + \dfrac {x^{r + 1} - 1} {r + 1}
      | c =
}}
{{eqn | r = H_{r + 1} + \sum_{k \mathop = 1}^{r + 1} \dbinom {r + 1} k \dfrac {\paren {x - 1}^k} k
      | c =
}}
{{end-eqn}}
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 1}: \ds \sum_{k \mathop = 1}^n \dfrac {x^k} k = H_n + \ds \sum_{k \mathop = 1}^n \dbinom n k \dfrac {\paren {x - 1}^k} k$