Problem on inequality with power sum Let $m$ and $n$ are positive integer

Can it be shown that, For every $m\ge 5$
$$\sum_{i=1}^ni^m-m^i>0\iff n=2,3,...,m$$

Example: let $m=5$, choose any $n$ between $2$ to $5$, now let $n=2$ then $\sum_{i=1}^2i^5-5^i=(1^5-5^1)+(2^5-5^2)=-4+7=3>0$
More on observation, $\sum_{i=1}^ni^m-m^i=0$ holds only for $(m,n)=\{(1,1),(2,3),(2,4)\}$
Score Pari/GP
for(m=5,50,for(n=1,50,if(sum(i=1,n,i^m-m^i)>0,print([m,n]))))

 A: Lemma 1: $2^m> m^2+m-1, \forall m>5$.
When $m=5, 2^5=32>29=5^2+5-1$. If $2^m>m^2+m-1$ then $$2^{m+1}-[(m+1)^2+(m+1)-1]\\>2(m^2+m-1)-(m+1)^2-(m+1)+1 = m(m-1)-3 >0.$$
Lemma 2: $i^j < j^i, \forall i>j>e.$
We check the first derivative of the function $f(x)=x^\frac 1x$:
$$f'(x)=x^{\frac 1x -2}(1-\ln x) < 0, \forall x>e.$$
First we prove $\Leftarrow$.
From the lemmas
$$\sum_{i=1}^n (i^m-m^i)=1+2^m-m-m^2+\sum_{i=3}^n(i^m-m^i)>0.$$
Next we prove $\implies$.
If $n=1$, it's trivial.
If $n>m$,
$$\sum_{i=1}^n (i^m-m^i)=\sum_{i=1}^{m+1} (i^m-m^i) + \sum_{i>m+1} (i^m-m^i) \le \sum_{i=1}^{m+1} (i^m-m^i) \text{ via Lemma 2}$$
Note that $e^{-\frac km} \ge 1-\frac km \implies \left(1-\frac km \right)^m \le e^{-k}, k=0, 1, \cdots, m-1$
Hence
$$
\sum_{i=1}^{m+1} (i^m-m^i) = \sum_{i=1}^{m+1} i^m - \sum_{i=1}^{m+1} m^i = m^m\left(\left(1+\frac 1m\right)^m + \sum_{k=0}^{m-1} \left(1-\frac km\right)^m \right) - \frac{m^{m+2} - m}{m-1}\\
< m^m \left(e+\sum_{k=0}^\infty e^{-k} - \frac{m^2-m^{1-m}}{m-1}\right) < m^m \left(e+\frac{1}{1-e^{-1}} - m\right) \approx m^m(4.3003-m) <0.\blacksquare
$$
