The platform still wants to attract content creators, and is currently working on a new subscription system… and paid.
As on Patreon and OnlyFans, Internet users will soon be subscribing to their favorite influencers and influencers on Instagram. For several years now, the platform has openly flirt with influencers. In addition to increasing the tools to enhance creation, Mark Zuckerberg’s social network recently mentioned the possibility of developing paid subscription systems, in order to allow followers “Premium” to benefit from exclusive content. An idea which is reminiscent of the Super Follows of Twitter, and which could very soon see the light of day.
According to the specialized site TechCrunch, which cites sources outside the company, Instagram would have added to its American application several new integrated purchasing options, in order to promote these famous subscriptions. If we trust these novelties, we can therefore imagine that the idea of a premium subscription on Instagram is about to be deployed internationally.
According to the company Sensor Tower, two purchase options would have been added since the beginning of November on the American App Store: a first for $ 0.99, and a second for $ 4.99. Until now, the only in-app payments offered were those relating to badges, gifts offered to content creators in the form of stickers during their lives. The arrival of paid subscriptions could therefore expand the offer of the platform, which has been looking for a long time how to keep its most influential creators.
However, it will probably be necessary to wait a little longer before taking advantage of these subscriptions. As the platform has become accustomed to testing its new features in a small area, it’s a safe bet that it will wait a few weeks before rolling out Premium subscriptions to the rest of the world. Until then, the company should lift the veil on the different types of subscriptions offered. Hopefully the exclusive content promised by the platform will be less subject to the censorship of its algorithm.