[Updated 4/22/25 to add n = 16 to 18, because I found a listing of the coefficients
from n = 1 to 18 from 01.18.76 printed by this BASIC program]
Example coefficients
$\displaystyle \sum_{i=1}^{n}i^{0}=n$
$\displaystyle \sum_{i=1}^{n}i^{1}=\frac{1}{2}n^{2}+\frac{1}{2}n$
$\displaystyle \sum_{i=1}^{n}i^{2}=\frac{1}{3}n^{3}+\frac{1}{2}n^{2}+\frac{1}{6}n$
$\displaystyle \sum_{i=1}^{n}i^{3}=\frac{1}{4}n^{4}+\frac{1}{2}n^{3}+\frac{1}{4}n^{2}$
$\displaystyle \sum_{i=1}^{n}i^{4}=\frac{1}{5}n^{5}+\frac{1}{2}n^{4}+\frac{1}{3}n^{3}-\frac{1}{30}n$
$\displaystyle \sum_{i=1}^{n}i^{5}=\frac{1}{6}n^{6}+\frac{1}{2}n^{5}+\frac{5}{12}n^{4}-\frac{1}{12}n^{2}$
$\displaystyle \sum_{i=1}^{n}i^{6}=\frac{1}{7}n^{7}+\frac{1}{2}n^{6}+\frac{1}{2}n^{5}-\frac{1}{6}n^{3}+\frac{1}{42}n$
$\displaystyle \sum_{i=1}^{n}i^{7}=\frac{1}{8}n^{8}+\frac{1}{2}n^{7}+\frac{7}{12}n^{6}-\frac{7}{24}n^{4}+\frac{1}{12}n^{2}$
$\displaystyle \sum_{i=1}^{n}i^{8}=\frac{1}{9}n^{9}+\frac{1}{2}n^{8}+\frac{2}{3}n^{7}-\frac{7}{15}n^{5}+\frac{2}{9}n^{3}-\frac{1}{30}n$
$\displaystyle \sum_{i=1}^{n}i^{9}=\frac{1}{10}n^{10}+\frac{1}{2}n^{9}+\frac{3}{4}n^{8}-\frac{7}{10}n^{6}+\frac{1}{2}n^{4}-\frac{3}{20}n^{2}$
$\displaystyle \sum_{i=1}^{n}i^{10}=\frac{1}{11}n^{11}+\frac{1}{2}n^{10}+\frac{5}{6}n^{9}-n^{7}+n^{5}-\frac{1}{2}n^{3}+\frac{5}{66}n$
$\displaystyle \sum_{i=1}^{n}i^{11}=\frac{1}{12}n^{12}+\frac{1}{2}n^{11}+\frac{11}{12}n^{10}-\frac{11}{8}n^{8}+\frac{11}{6}n^{6}-\frac{11}{8}n^{4}+\frac{5}{12}n^{2}$
$\displaystyle \sum_{i=1}^{n}i^{12}=\frac{1}{13}n^{13}+\frac{1}{2}n^{12}+n^{11}-\frac{11}{6}n^{9}+\frac{22}{7}n^{7}-\frac{33}{10}n^{5}+\frac{5}{3}n^{3}-\frac{691}{2730}n$
$\displaystyle \sum_{i=1}^{n}i^{13}=\frac{1}{14}n^{14}+\frac{1}{2}n^{13}+\frac{13}{12}n^{12}-\frac{143}{60}n^{10}+\frac{143}{28}n^{8}-\frac{143}{20}n^{6}+\frac{65}{12}n^{4}-\frac{691}{420}n^{2}$
$\displaystyle \sum_{i=1}^{n}i^{14}=\frac{1}{15}n^{15}+\frac{1}{2}n^{14}+\frac{7}{6}n^{13}-\frac{91}{30}n^{11}+\frac{143}{18}n^{9}-\frac{143}{10}n^{7}+\frac{91}{6}n^{5}-\frac{691}{90}n^{3}+\frac{7}{6}n$
$\displaystyle \sum_{i=1}^{n}i^{15}=\frac{1}{16}n^{16}+\frac{1}{2}n^{15}+\frac{5}{4}n^{14}-\frac{91}{24}n^{12}+\frac{143}{12}n^{10}-\frac{429}{16}n^{8}+\frac{455}{12}n^{6}-\frac{691}{24}n^{4}+\frac{35}{4}n^{2}$
$\displaystyle \sum_{i=1}^{n}i^{16}=\frac{1}{17}n^{17}+\frac{1}{2}n^{16}+\frac{4}{3}n^{15}-\frac{14}{3}n^{13}+\frac{52}{3}n^{11}-\frac{143}{3}n^{9}+\frac{260}{3}n^{7}-\frac{1382}{15}n^{5}+\frac{140}{3}n^{3}-\frac{3617}{510}n$
$\displaystyle \sum_{i=1}^{n}i^{17}=\frac{1}{18}n^{18}+\frac{1}{2}n^{17}+\frac{17}{12}n^{16}-\frac{17}{3}n^{14}+\frac{221}{9}n^{12}-\frac{2431}{30}n^{10}+\frac{1105}{6}n^{8}-\frac{11747}{45}n^{6}+\frac{595}{3}n^{4}-\frac{3617}{60}n^{2}$
$\displaystyle \sum_{i=1}^{n}i^{18}=\frac{1}{19}n^{19}+\frac{1}{2}n^{18}+\frac{3}{2}n^{17}-\frac{34}{5}n^{15}+34n^{13}-\frac{663}{5}n^{11}+\frac{1105}{3}n^{9}-\frac{23494}{35}n^{7}+714n^{5}-\frac{3617}{10}n^{3}+\frac{43867}{798}n$
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