Consider numbers of the form of $k^m(k^i+j)$ for $i\in\{1,\ldots, n-k+1\}$, $j\in \{1,\ldots,k-1\}$ and $m\geq 0$. Note that for such $i$ and $j$ in these intervals, we have $i+j\le n$. Based on $n-(i+j)$ is an odd number or an even number, we have two cases. $\bullet$ If $n-(i+j)$ is odd, there is some integer $t\geq 0$ such that $$n-(i+j)=2t+1.$$In this case, the number $k^i+j$ can be represented using $n$ number of $k$'s as $$k^i+j=\underbrace{\left( k\times \cdots \times k\right )}_{i+t}\div\underbrace{(k\times\cdots\times k)}_{t}+\underbrace{(k+\cdots+k)}_j\div k.$$$\bullet$ If $n-(i+j)$ is even, there is some integer $t\geq 0$ such that $$n-(i+j)=2t.$$In this case, the number $k(k^i+j)$ can be shown using $n$ number of $k$'s as $$k(k^i+j)=k\times(\underbrace{\left( k\times \cdots \times k\right )}_{i+t-1}\div\underbrace{(k\times\cdots\times k)}_{t-1}+\underbrace{(k+\cdots+k)}_j\div k).$$(In case when $t=0$, we have $t-1=-1$, so in the above equation, instead of multiplying the parentheses to $-1$ $k$'s, we divide it by one $k$.) So in both cases, there is $m\ge0$ such that $k^m(k^i+j)$ can be be represented using $n$ number of $k$'s. Set $m$ to be the largest number such that $k^m(k^i+j)$ can be represented using only $n$ number of $k$'s (for fixed $i$ and $j$). There are a total of $(n-k+1)(k-1)$ numbers created with given properties. We prove that they are all distinct numbers and also cannot be represented using $n-1$ number of $k$'s. If some number $u=k^m(k^i+j)$ can be shown using $n-1$ number of $k$'s, then clearly $k\times u=k^{m+1}(k^i+j)$ has a representation, using $n$ number of $k$'s, therefore $m$ cannot be the largest number such that $k^m(k^i+j)$ has a representation using $n$ number of $k$'s, so we conclude that it is impossible to reach these numbers with only $n-1$ number of $k$'s. Finally, if for two such triples $(i_1,j_1,m_1)\neq (i_2,j_2, m_2)$ with $m_1\geq m_2$, we have $$k^{m_{1}}\left ( k^{i_{1}}+j_{1} \right )=k^{m_{2}}\left(k^{i_{2}}+j_{2}\right),$$then $$k^{m_{1}-m_{2}}\left ( k^{i_{1}}+j_{1} \right )=k^{i_{2}}+j_{2}.$$But if $m_1-m_2>0$, the above equation implies $k\mid j_2$, which is impossible because $j_2<k$. Therefore $m_1=m_2$ and so $ k^{i_{1}}+j_{1} =k^{i_{2}}+j_{2}.$ Without loss of generality, we assume $i_1\ge i_2$. So $$k^{i_{2}}\left ( k^{i_{1}-i_{2}}-1 \right )=j_2-j_1.$$Again, if $i_1>i_2$, we get $k\mid j_2-j_1$. Note that $j_1,j_2\in\{1,\ldots,k-1\}$, therefore $-k<\left | j_1-j_2 \right |<k$, so we must have $j_1-j_2=0$ and thus $k^{i_1}=k^{i_2}$ which means $i_1=i_2$, that is impossible in this case be cause we assumed that $i_1>i_2$. The only possible case is when $i_1=i_2$ which leads to $j_1=j_2$. Therefore $(i_1,j_1,m_1)= (i_2,j_2, m_2)$, contradiction. So indeed, all of the created numbers are distinct and hence the claim of the problem. $\blacksquare$