| Election Name: | 1890 Idaho gubernatorial election |
| Country: | Idaho |
| Type: | Presidential |
| Ongoing: | no |
| Next Election: | 1892 Idaho gubernatorial election |
| Next Year: | 1892 |
| Election Date: | 1 October 1890 |
| Nominee1: | George L. Shoup |
| Party1: | Republican Party (United States) |
| Popular Vote1: | 10,262 |
| Percentage1: | 56.35% |
| Nominee2: | Benjamin Wilson |
| Party2: | Democratic Party (United States) |
| Popular Vote2: | 7,948 |
| Percentage2: | 43.65% |
| Map Size: | 160px |
| Governor | |
| Before Election: | George L. Shoup (Territorial) |
| Before Party: | Republican Party (United States) |
| After Election: | George L. Shoup |
| After Party: | Republican Party (United States) |
The 1890 Idaho gubernatorial election was held on 1 October 1890, in order to elect the first Governor of Idaho upon Idaho acquiring statehood in July 1890. Incumbent Republican Governor of the Idaho Territory George L. Shoup defeated Democratic nominee Benjamin Wilson.[1]
On election day, 1 October 1890, Republican nominee George L. Shoup won the election by a margin of 2,314 votes against his opponent Democratic nominee Benjamin Wilson, thereby retaining Republican control over the new office of Governor. Shoup was sworn in as the first Governor of the new state of Idaho on 8 December 1890.[2]