In trying to compute that $\pi_1(U(n))\cong \mathbb{Z}$, I came across answers that considered concepts such as Long Exact Sequences, which I have no knowledge of. Instead, I tried to show that $U(n) \simeq U(1)$ have the same homotopy type by considering the maps $det:U(n)\to U(1)$, which is the determinant map, and the inclusion $i:U(1)\to U(n)$ which sends the complex $z$, where $|z|=1$, to the matrix $\ diag(z, I_{n-1})$, i.e. the $n\times n$ identity matrix except the first entry is equal to $z$. It seems clear that $det\circ i = id_{U(1)}$, and if I knew $U(n)$ was path connected, then there is a path from any matrix $\ A\ $ to the matrix $\ i\circ det (A)$, which means there is a homotopy $\ i\circ det \simeq id_{U(n)}$. Since $U(1)\cong S^1$, this would show $\pi_1(U(n))\cong \pi_1(U(1))\cong \pi_1(S^1) \cong \mathbb{Z}$.
Does this method of computing $\pi_1(U(n))$ work? Or is there something that I did not consider that might invalidate this method?
EDIT: Thank you for your answers, I now see why $U(n)\not \simeq U(1)$. But then, what is the flaw in my proof idea?
Note that $SU(2)$ is isomorphic to the $3$-sphere $S^3$, and the inclusion $SU(2)\hookrightarrow U(2)$ gives rise to a non-trivial element of $\pi_3(U(2))$. On the other hand $\pi_3(U(1))=0$, so $U(1)$ and $U(2)$ cannot be homotopy equivalent.
I think the simplest way to compute $\pi_1(U(n))$ is to use the long exact sequence in homotopy groups induced by the short exact sequence $1\to SU(n)\to U(n)\to U(1)\to 1$, which is probably what the answers you have seen uses.